In the previous chapter, we studied how to combine primitive propositions with the operators, ∧, ∨, —>, and ¬. When we wrote propositions like (p ∧ q) —> r, we pretended that p, q, and r stood for complete, primitive statements like ``It is raining'' or ``x + 1 > 0''. We did not try to disassemble p, q, and r.
Now it is time to decompose and analyze primitive propositions in terms of their ``verbs'' (called predicates) and their ``nouns'' (called individuals). This leads to predicate logic.
First, some background: When we study a particular ``universe'' or ``domain'' consisting of ``individuals,'' we make assertions (propositions) about the individuals in the domain. Example domains are: the domain of all animals, the domain of U.S. Presidents, the domain of days-of-the-week, the domain of crayon colors, the domain of integers, the domain of strings, etc. We assemble propositions by using the individuals in the domain along with some predicates. For example, for the domain of integers, we use predicates like == and >, like this: 3 > 5, 2 * x == y + 1, etc. (Here, 3 and 5 are individuals, and x and y are names of individuals.) As these examples show, we might also use functions, like * and +, to compute new individuals.
For nonnumeric domains like humans, animals, and objects, predicates are written in a function-call style, like this: hasFourLegs(_), isTheMotherOf(_,_), isHuman(_), isOlderThan(_,_), etc. So, if Lassie is an individual animal, we write hasFourLegs(Lassie) to make the proposition, ``Lassie has four legs.'' Another example is isOlderThan(GeorgeWashington, AbrahamLincoln), which uses the individuals GeorgeWashington and AbrahamLincoln.
Predicate logic has two important new operators that let us write stronger propositions than what we can do with mere predicates. These operators are called quantifiers. The quantifers are ``for all'' (∀), and ''exists'' (∃). In this chapter, we will learn to use the quantifiers to reason about data structures.
The ∀ quantifier helps us write propositions about all the individuals in a domain. Say we consider the domain of animals. The sentence, ``All humans are mortal'' is written like this:
An arithmetic example looks like this: for the domain of ints, ``every value is less-than-or-equal to its square'':
The previous statement is often written in a ``shorthand'' like this:
The ∃ quantifier helps us write propositions about specific individuals in a domain, where the name of the individual is unimportant or unknown. For example, we can say that Lassie has a mother like this:
∃x (x * x = x) ∃y (y + 2 = 9) ∀x (x > 1) —> (∃y (y > 0 and y + 1 = x))For array r, we can say that r holds at least one negative int like this:
∃i (i >= 0 ∧ i < len(r) ∧ r[i] < 0)(The shorthand version is ∃ 0 <= i < len(r), r[i] < 0.)
Lots more examples will follow.
With the new format of primitive propositions, we can write proofs like before:
=================================================== isHuman(Socrates) —> isMortal(Socrates), isHuman(Socrates) |− isMortal(Socrates) ∧ isHuman(Socrates) 1. isHuman(Socrates) —> isMortal(Socrates) premise 2. isHuman(Socrates) premise 3. isMortal(Socrates) —>e 1,2 4. isMortal(Socrates) ∧ isHuman(Socrates) ∧i 3,2 ===================================================But more importantly, we will learn to prove claims like this:
∀x(isHuman(x) —> isMortal(x)), isHuman(Socrates) |− isMortal(Socrates)
∀x (isHuman(x) —> isMortal(x)), isHuman(Socrates) |− isMortal(Socrates)Clearly, we require a kind of matching/binding rule to prove that the human individual, Socrates, is mortal. The rule is ∀e (``all elimination''):
=================================================== 1. ∀x (isHuman(x) —> isMortal(x)) premise 2. isHuman(Socrates) premise 3. isHuman(Socrates) —> isMortal(Socrates) ∀e 1 4. isMortal(Socrates) —>e 3,2 ===================================================Line 3 shows that the claim on Line 1, which holds for all individuals in the domain, can apply specifically to Socrates, an individual member of the domain. We use the new knowledge on Line 3 to reach the conclusion on Line 4.
∀e tailors a general claim, prefixed by ∀x, to any individual element (who replaces the x). We see this in Line 3 above. Here is the rule's schematic:
=================================================== ∀x P_x ∀e: ------------ P_v that is, [v/x]P_x, where v is an individual in the domain ===================================================(Here, P_x stands for a proposition that contains some occurrences of x. Recall that [v/x]P_x is ``substitution notation'': substitute v for occurrences of x in P_x.) For example, from the premise, ∀i (i + 1 > i), we apply ∀e to deduce [3/i](i + 1 > i), that is, 3 + 1 > 3.
The other deduction rule, ∀i (``all-introduction''), deduces propositions that are prefixed by ∀. Here is a motivating example, in the domain of integers:
∀n((n + 1) > n), ∀n(n > (n - 1)) |− ∀n((n + 1) > n ∧ n > (n - 1))That is, we wish to prove that for every possible int, the int is smaller than its successor and larger than its predecessor. How do we do this?
Clearly, we will not inspect all of ..., -2, -1, 0, 1, 2, ... and verify that (-2 + 1) > -2 ∧ -2 < (-2 - 1), (-1 + 1) > -1 ∧ -1 < (-1 - 1), (0 + 1) > 0 ∧ 0 < (0 - 1), etc.! Instead, we write a single, generic, general-purpose argument --- a ``case analysis'' --- that applies to whichever, arbitrary int we would ever consider. Let a stand for the arbitrary int we will discuss. The case analysis appears in the proof like this:
=================================================== ∀n ((n + 1) > n), ∀n (n > (n - 1)) |− ∀n ((n + 1) > n ∧ n > (n - 1)) 1. ∀n ((n + 1) > n) premise 2. ∀n (n > (n - 1)) premise ... 3. a ... 4. (a + 1) > a ∀e 1 ... 5. a > (a - 1) ∀e 2 ... 6. (a + 1) > a ∧ a > (a - 1) ∧i 4,5 7. ∀n ((n + 1) > n ∧ n > (n - 1)) ∀i 3-6 ===================================================Lines 3-6 are the generic argument: let a be the arbitrary/anybody integer we discuss. By Lines 1 and 2, we must have that (a + 1) > a and that a > (a - 1). Line 6 uses ∧i to show a has the property (a + 1) > a ∧ a > (a - 1).
Since the argument in Lines 3-6 is not specific to any specific integer, we can use the argument on all the individual integers --- that is, we can substitute -2 for a and the argument holds; we can substitute -1 for a and the argument holds; we can substitute 0 for a and the argument holds; and so on!
Line 7 is justified by the new deduction rule, ∀i, which asserts that the generic case analysis in Lines 3-6 applies to all the individual integers. Here is the rule's schematic:
=================================================== ... a (a must be a brand new name) ... P_a ∀i: ------------ ∀x P_x (That is, P_x is [x/a]P_a. Thus, a _does not appear_ in P_x, and every premise and assumption visible to ∀x P_x _does not mention_ a) ===================================================
To repeat this important idea: The rule says, to prove a claim of form, ∀x P_x, we undertake a case analysis: we prove property P_a for an arbitrary member, a, of domain D. (Call the element, ``Mister a'' --- Mister arbitrary --- Mister anybody --- Mister anonymous). Since Mister a is a complete unknown, it stands for ``everyone'' in doman D. We know that we can substitute whichever domain element, d, from domain D we want into the proof and we get a proof of P_d. In this way, we have proofs of P for all elements of domain D.
Here is the same idea, used in a proof about a domain of people: ''Everyone is healthy; everyone is happy. Therefore, everyone is both healthy and happy'':
∀x isHealthy(x), ∀y isHappy(y) |− ∀z(isHealthy(z) ∧ isHappy(z))
=================================================== 1. ∀x isHealthy(x) premise 2. ∀y isHappy(y) premise ... 3. a ... 4. isHealthy(a) ∀e 1 ... 5. isHappy(a) ∀e 2 ... 6. isHealthy(a) ∧ isHappy(a) ∧i 4,5 7. ∀z(isHealthy(z) ∧ isHappy(z)) ∀i 3-6 ===================================================
Say that we have a domain of living beings. This next example requires nested cases:
=================================================== ∀x (isHuman(x) —> isMortal(x)), ∀y (isMortal(y) —> hasSoul(y)) |− ∀x (isHuman(x) —> hasSoul(x)) 1. ∀x (isHuman(x) —> isMortal(x)) premise 2. ∀y (isMortal(y) —> hasSoul(y)) premise ... 3. a ... ... 4. isHuman(a) assumption ... ... 5. isHuman(a) —> isMortal(a) ∀e 1 ... ... 6. isMortal(a) —>e 4,3 ... ... 7. isMortal(a) —> hasSoul(a) ∀e 2 ... ... 8. hasSoul(a) —>e 6,5 ... 9. isHuman(a) —> hasSoul(a) —>i 4-8 10. ∀x (isHuman(x) —> hasSoul(x)) ∀i 3-9 ===================================================Line 3 states that we use a to stand for an arbitrary individual of the domain. Line 4 starts a nested case, which assumes a is human. Then we can prove that a has a soul, hence by —>i, isHuman(a) —> hasSoul(a). Since the outer case is stated in terms of the arbitrary, anonymous individual, a, we can finish the proof on Line 10 by ∀i.
Here is a last, important example. Let the domain be the members of one family. We can prove this truism:
=================================================== ∀x (healthy(x) —> happy(x)) |− (∀y healthy(y)) —> (∀x happy(x)) 1. ∀x healthy(x) —> happy(x) premise ... 2. ∀y healthy(y) assumption ... ... 3. a ... ... 4. healthy(a) ∀e 4 ... ... 5. healthy(a) —> happy(a) ∀e 1 ... ... 6. happy(a) —>e 5,4 ... 7. ∀ x happy(x) ∀i 3-6 8. (∀y healthy(y)) —> (∀x happy(x)) —>i 2-7 ===================================================We commence by assuming all the family is healthy (Line 2). Then, we consider an arbitrary/anonymous family member, a, and show that healthy(a) is a fact (from Line 2). Then we deduce happy(a). Since a stands for anyone/everyone in the family, we use foralli to conclude on Line 7 that all family members are happy. Line 8 finishes.
Consider the converse claim; is it valid?
Let's try to prove the dubious claim and see where we get stuck:
=================================================== (∀y healthy(y)) —> (∀x happy(x)) |− ∀x (healthy(x) —> happy(x)) 1. (∀y healthy(y)) —> (∀x happy(x)) premise ... 2. a assumption ... ... 3. healthy(a) assumption WE ARE TRYING TO PROVE happy(a)?! 4. ∀y healthy(y) ∀i 2-3?? NO --- WE ARE TRYING TO FINISH THE OUTER BLOCK BEFORE THE INNER ONE IS FINISHED! ===================================================No matter how you might try, you will see that the ``block structure'' of the proofs warns us when we are making invalid deductions. It is impossible to prove this claim.
We state some standard exercises with ∀, where the domains and predicates are unimportant:
=================================================== ∀x F(x) |− ∀y F(y) 1. ∀x F(x) premise ... 2. a ... 3. F(a) ∀e 1 4. ∀y F(y) ∀i 2-3 ===================================================
=================================================== ∀z (F(z) ∧ G(z) |− (∀x F(x)) ∧ (∀y G(y)) 1. ∀z (F(z) ∧ G(z) premise ... 2. a ... 3. F(a) ∧ G(z) ∀e 1 ... 4. F(a) ∧e1 3 5. ∀x F(x) ∀i 2-4 ... 6. b ... 7. F(b) ∧ G(b) ∀e 1 ... 8. G(b) ∧e2 7 9. ∀y F(y) ∀i 6-8 10. (∀x F(x)) ∧ (∀y G(y)) ∧i 5,9 ===================================================The earlier example about healthy and happy families illustrates an important structural relationship between ∀ and —>:
∀x (F(x) —> G(x)) |− (∀x F(x)) —> (∀x G(x))can be proved, but the converse cannot.
This last one is reasonable but the proof is a bit tricky because of the nested subproofs:
=================================================== ∀x ∀y F(x,y) |− ∀y ∀x F(x,y) 1. ∀x ∀y F(x,y) premise ... 2. b ... ... 3. a ... ... 4. ∀y F(a,y) ∀e 1 ... ... 5. F(a,b) ∀e 4 ... 6. ∀x F(x,y) ∀i 3-5 7. ∀y ∀x F(x,y) ∀i 2-6 ===================================================
1. Premises premise ... i. a (fill in) ... j. P_a k. ∀x P_x ∀i i-jThis tactic was applied in Lines 2-7 of the previous (correct) example proof.
1. Premises premise 2. ∀x P_x premise . . . j. P_i ∀e 2 (fill in) k. QThis tactic should be used only when it is clear that the new fact makes a significant step forwards to finishing the proof. Steps 4 and 5 of the previous (correct) example proof used this tactic.
Approach 1: use conjunctions for a finite domain
This makes ∀x P_x just an abbreviation itself of this much-longer assertion:
P_e0 ∧ P_e1 ∧ ... ∧ P_ekFor example, when the domain is the days of the week, the assertion, ∀d isBurgerKingDay(d), abbreviates
isBurgerKingDay(sun) ∧ isBurgerKingDay(mon) ∧ isBurgerKingDay(tues) ∧ ... ∧ isBurgerKingDay(sat)To prove such a ∀x P_x for a finite domain D, we must prove P_ei, for each and every ei in D.
We can use this approach when we are analyzing all the elements of a finite-length array. Say that array r has length 4. We can say that the domain of its indexes is {0, 1, 2, 3}. So, if we wish to prove that ∀ 0 <= i < 4, r[i] > 0, we need only prove that r[0] > 0 ∧ r[1] > 0 ∧ r[2] > 0 ∧ r[3] > 0.
Approach 2: for the domain of nonnegative ints, use mathematical induction
There is a variation of mathematical induction that we use when proving loop invariants of the form, ∀ 0 <= k < count, P_k:
... count = 0 { (a) ∀ 0 <= k < count, P_k } while B { (b) invariant ∀ 0 <= k < count, P_k } ... { retain: ∀ 0 <= k < count, P_k prove: P_k implies: ∀ 0 <= k < count + 1, P_k } count = count + 1 { (c) ∀ 0 <= k < count, P_k }
∀ 0 <= k < 0, P_kThis assertion is true because it defines an empty range of integers --- there are no elements in the domain defined by {k:int | 0 <= k ∧ k < 0}. Hence, for all the elements, k, in the empty domain, we have proved P_k(!) This proves the Basis case.
P_countThat is, P holds for the value, count. Using ∧i we get
∀ 0 <= k < count, P_k ∧ P_countThis can be understood as
(P_0 ∧ P_1 ∧ P_2 ∧ ... ∧ P_count-1) ∧ P_countBased on what we read in Approach 1 earlier, we combine the facts into this one:
∀ 0 <= k < count + 1, P_kSince the loop's body ends with the assignment, count = count + 1, we recover the loop invariant at point (c). This is the proof of the induction case.
We will see several uses of this approach in the next Section.
Approach 3: for any domain, finite or infinite whatsoever, use the ∀i-law
Finally, we might be using a large domain that is not as organized as the nonnegatives, 0,1,2,.... Maybe the domain is the domain of all humans or all the citizens of Peru or or the members of the Republican party or all the objects on Planet Earth. How can we prove ∀ x P_x for such huge collections?
To prove a claim of form, ∀x P_x, for an arbitrary domain, we undertake a kind of case analysis: we prove property P_a for an arbitrary member, a, of domain D. (Call the element, ``Mister a'' --- Mister arbitrary --- Mister anybody --- Mister anonymous). Since Mister a is a complete unknown, it stands for ``everyone'' in doman D. We know that we can substitute whichever domain element, d from domain D, we want into the proof and we get a proof of P_d. In this way, we have proofs of P for all elements of domain D.
This is the idea behind the ∀i-rule.
=================================================== def fac(n) : { pre n >= 0 post ans == n! return ans } ===================================================We saw in the previous chapter that the pre- and post-condition can be combined into this compound proposition,
(n >= 0) —> (fac(n) == n!)which describes fac's behavior in terms of name n.
But n is a parameter name that is internal to fac's code. A more proper specification, one that makes sense to fac's caller, is
∀n ((n >= 0) —> (fac(n) == n!))That is, ``for all possible int arguments, if the argument is nonnegative, then fac computes the argument's factorial.''
We use this logical propery when we call the function. Here is the proof of it:
=================================================== x = fac(6) { 1. ∀n((n >= 0) —> (fac(n) == n!)) premise (about fac ) 2. 6 >= 0 algebra 3. (6 >= 0) —> (fac(6) == 6!) ∀e 1 4. fac(6) == 6! —>e 3,2 5. x == fac(6) premise (the assign law) 6. x == 6! subst 4,5 } ===================================================∀e applies the function's logical property to its argument. The function-call law we learned in Chapter 3 hid the ∀ --- we weren't ready for it yet! But the universal quantifier is implicit in the description of every function we call.
When we wrote the coding of fac, we also built a proof that fac computes and returns n!, for input parameter name, n. We didn't know if n will equal 1 or 9 or 99999 --- we just call it n and work with the name. This is just a ``Mr. anybody'', exactly as we have been using in our case analyses that finish with ∀i. The rule for function building hides the use of ∀i --- we were not ready for it in Chapter 3. But writing a function is the same thing as writing a proof that finishes with ∀i.
We use arrays (lists) in the examples in this chapter. First, recall these Python operators for arrays:
=================================================== a = [2, 3, 5, 7] print a # prints [2, 3, 5, 7] a.append(11) print a # prints [2, 3, 5, 7, 11] ===================================================
=================================================== c = [2, 3, 5, 7, 11, 13, 17, 19] e = c[:6] print e # prints [2, 3, 5, 7, 11, 13] f = c[:0] print f # prints [] print c # prints [2, 3, 5, 7, 11, 13, 17, 19] ===================================================
=================================================== c = [2, 3, 5, 7, 11, 13, 17, 19] g = c[4:] print g # prints [11, 13, 17, 19] h = c[:8] print h # prints [] print c # prints [2, 3, 5, 7, 11, 13, 17, 19] ===================================================
Here is an example; it is math-induction-like (``Approach 2'' mentioned earlier). We outline below how a procedure resets all the elements of an array (list) to zeros:
=================================================== def zeroOut(a) : { pre isIntArray(a) post ∀ 0 <= i < len(a), a[i] == 0 } j = 0 while j != len(a) : { invariant ∀ 0 <= i < j, a[i] == 0 } a[j] = 0 { assert ∀ 0 <= i < j, a[i] == 0 ∧ a[j] = 0 therefore, ∀ 0 <= i < j+1, a[i] == 0 (*) } j = j + 1 { assert ∀ 0 <= i < j, a[i] == 0 } #END LOOP { assert j == len(a) ∧ (∀ 0 <= i < len(a), a[i] == 0) therefore, ∀ 0 <= i < len(a), a[i] == 0 } ===================================================We state that the range of elements from 0 up to (and not including) j are reset to 0 by stating
∀ 0 <= i < j, a[i] == 0This loop invariant leads to the goal as j counts through the range of 0 up to the length of array a. At the point marked (*), there is an informal use of ∀i.
Here is a second, similar example:
=================================================== def doubleArray(a) : """doubleArray builds a new array that holds array a's values *2""" { pre: isIntArray(a) post: isIntArray(answer) ∧ len(answer) == len(a) ∧ ∀ 0 <= i < len(a), answer[i] == 2 * a[i] } index = 0 answer = [] while index != len(a) : { invariant isIntArray(answer) ∧ len(answer) == index ∧ ∀ 0 <= i < index, answer[i] == 2 * a[i] } { assert: index != len(a) ∧ invariant } answer.append([a[index]*2) { assert: invariant ∧ answer[index] == 2 * a[index] implies: ∀ 0 <= i < index+1, answer[i] == 2 * a[i] } # (see Approach 2) index = index + 1 { assert: invariant } { assert: index == len(a) ∧ invariant implies: isIntArray(answer) ∧ len(answer) == len(a) implies: ∀ 0 <= i < len(a), answer[i] == 2 * a[i] } return answer ===================================================Notice how the postcondition notes that the answer array is the same length as the parameter array. This prevents the function's code from misbehaving and adding junk to the end of the answer array.
See the Case Studies for more examples.
There is a mouse in the house: ∃m (isMouse(m) ∧ inHouse(m)) (We don't care about the mouse's name.) Someone ate my cookie: ∃x ateMyCookie(x) There is a number that equals its own square: ∃n n == n*n For every int, there is an int that is smaller: ∀x ∃y y < xIf we have a fact about an individual in a domain, we can use the fact to deduce a fact that begins with an existential quantifier. For example, if we know that
isHuman(Socrates) ∧ isMortal(Socrates)surely we can conclude that
∃h (isHuman(h) ∧ isMortal(h))that is, ``there is someone who is human and mortal.'' The identity of the human is no longer important to us. In the next section, we see that the ∃i-rule makes such deductions.
Often ∃ is used to ``hide'' secret information. Consider these Pat Sajack musings from a typical game of Wheel of Fortune:
∃s (isSquare(s) ∧ isCovered(s) ∧ (∃c isVowel(c) ∧ holds(s,c)))This statement hides the specific square and letter that Pat is thinking about.
What can a game player do with Pat's uttered statement? A player might deduce these useful facts:
P_d where d is an individual in the domain D ∃i: ------------ ∃x P_xThe ∃i rule says, if we locate an individual d (a ``witness'', as it is called by logicians) that makes P true, then surely we can say there exists someone that has P and hide the identity of the individual/witness.
The rule was used in the previous section in a tiny example:
=================================================== isHuman(Socrates), isMortal(Socrates) |− ∃h (isHuman(h) ∧ isMortal(h)) 1. isHuman(Socrates) premise 2. isMortal(Socrates) premise 3. isHuman(Socrates) ∧ isMortal(Socrates) ∧i 1,2 4. ∃h (isHuman(h) ∧ isMortal(h)) ∃i 3 ===================================================Since Socrates is an individual that is both human and mortal, we deduce Line 3. Line 4 ``hides'' Socrates's name.
Let's do a Wheel-Of-Fortune example: Pat Sajak uses two premises and the ∃i rule to deduce a new conclusion:
=================================================== isVowel('E'), holds(Square14,'E') |− ∃c(isVowel(c) ∧ ∃s holds(s,c)) 1. isVowel('E') premise 2. holds(Square14,'E') premise 3. ∃s holds(s,'E') ∃i 2 4. isVowel('E') ∧ ∃s holds(s,'E') ∧i 1,3 5. ∃c(isVowel(c) ∧ ∃s holds(s,c)) ∃i 4 ===================================================Line 3 hides the number of the square (``there is a square that holds 'E' ''), and Line 5 hides the 'E' (``there is a letter that is a vowel and there is a square that holds the letter'').
From the same two premises we can also prove this:
=================================================== 1. isVowel('E') premise 2. holds(Square14,'E') premise 3. isVowel('E') ∧ holds(Square14,'E') ∧i 1,3 4. ∃s(isVowel('E') ∧ holds(s,'E')) ∃i 3 5. ∃c∃s(isVowel(s) ∧ holds(s,c)) ∃i 4 ===================================================This reads, ``there are a letter and square such that the letter is a vowel and the square holds the letter.'' The proposition differs slightly from the previous one, but the two seem to have identical information content. (When we learn the ∃e-rule, we can prove the two conclusions have identical content.)
Here is a quick example (in the universe of things on planet Earth), to get our bearings:
All humans are mortal Someone is human Therefore, someone is mortalWe don't know the name of the individual human, but it does not matter --- we can still conclude someone is mortal. The steps we will take go like this:
Say we have a premise of the form, ∃x P_x. Since we do not know the name of the individual ``hidden'' behind the ∃x, we make up a name for it, say a, and discuss what must follow from the assumption that P_a holds true:
... a P_a assumption (where a is a new, fresh name) ∃x P_x ... Q ∃e: ----------------------- (a MUST NOT appear in Q) QThat is, if we can deduce Q from P_a, and we do not mention a within Q, then it means Q can be deduced no matter what name the hidden individual has. So, Q follows from ∃x P_x.
We can work the previous example, with ∃e:
All humans are mortal Someone is human Therefore, someone is mortalWe make up the name, a, for the individual whose name we do not know, and do a case analysis:
=================================================== ∀h(isHuman(h) —> isMortal(h)), ∃x isHuman(x) |− ∃y isMortal(y) 1. ∀h(isHuman(h) —> isMortal(h)) premise 2. ∃x isHuman(x) premise ... 3. a isHuman(a) assumption ... 4. isHuman(a) —> isMortal(a) ∀e 1 ... 5. isMortal(a) —>e 4,3 ... 6. ∃y isMortal(y) ∃i 5 7. ∃y isMortal(y) ∃e 2, 3-6 ===================================================Line 3 proposes the name a and the assumption that isHuman(a). The case analysis leads to Line 6, which says that someone is mortal. (We never learned the individual's name!) Since Line 6 does not explicitly mention the made-up name, a, we use Line 7 to repeat Line 6 --- without knowing the name of the individual ``hiding'' inside Line 2, we made a case analysis in Lines 3-6 that prove the result, anyway. This is how ∃e works.
To repeat: The ∃e rule describes how to discuss an anonymous individual (a witness) without knowing/revealing its identity: Assume the witness's name is Mister a (``Mister Anonymous'') and that Mister a makes P true. Then, we deduce some fact, Q, that holds even though we don't know who is Mister a. The restriction on the ∃e rule (Q cannot mention a) enforces that we have no extra information about the identity of Mister a --- the name a must not leave the subproof.
Here is a Wheel-of-Fortune example that uses ∃e:
=================================================== ∃c (isVowel(c) ∧ ∃s holds(s,c)) |− ∃y isVowel(y) 1. ∃c (isVowel(c) ∧ ∃s holds(s,c)) premise ... 2. a isVowel(a) ∧ ∃s holds(s,a) assumption ... 3. isVowel(a) ∧e1 2 ... 4. ∃y isVowel(y) ∃i 3 5. ∃y ∃y isVowel(y) ∃e 1,2-4 ===================================================We do not know the identity of the vowel held in an unknown square, but this does not prevent us from concluding that there is a vowel.
=================================================== ∃x P(x) |− ∃y P(y) 1. ∃x P(x) premise ... 2. a P(a) assumption ... 3. ∃y P(y) ∃i 2 4. ∃y P(y) ∃e 1,2-3 ===================================================
=================================================== ∃x(F(x) ∧ G(x)) |− (∃y F(y)) ∧ (∃z G(z)) 1. ∃x(F(x) ∧ G(x)) premise ... 2. a F(a) ∧ G(a) assumption ... 3. F(a) ∧e1 2 ... 4. ∃y F(y) ∃i 3 ... 5. G(a) ∧e2 2 ... 6. ∃z G(z) ∃i 5 ... 7. (∃y F(y)) ∧ (∃z G(z)) ∧i 4,6 8. (∃y F(y)) ∧ (∃z G(z)) ∃e 1, 2-7 ===================================================Notice that you cannot prove the converse: (∃y F(y)) ∧ (∃z G(z)) |− ∃x(F(x) ∧ G(x)). For example, say that the universe of discussion is the cast of Wheel of Fortune, and F == isMale and G == isFemale. Clearly, Pat Sajak is male and Vanna White is female, so (∃y F(y)) ∧ (∃z G(z)) is a true premise. But ∃x(F(x) ∧ G(x)) does not follow.
The following proof uses the ∨e-tactic --- a cases analysis. See the assumptions at lines 3 and 6, based on Line 2:
=================================================== ∃x (P(x) ∨ Q(x)) |− (∃x P(x)) ∨ (∃x Q(x)) 1. ∃x (P(x) ∨ Q(x)) premise ... 2. a P(a) ∨ Q(a) assumption ... ... 3. P(a) assumption ... ... 4. ∃x P(x) ∃i 3 ... ... 5. (∃x P(x)) ∨ (∃x Q(x)) ∨i1 4 ... ... 6. Q(a) assumption ... ... 7. ∃x Q(x) ∃i 6 ... ... 8. (∃x P(x)) ∨ (∃x Q(x)) ∨i2 7 ... 9. (∃x P(x)) ∨ (∃x Q(x)) ∨e 2,3-5,6-8 11. (∃x P(x)) ∨ (∃x Q(x)) ∃e 1,2-9 ===================================================As an exercise, prove the converse of the above: (∃x P(x)) ∨ (∃x Q(x)) |− ∃x (P(x) ∨ Q(x)).
We finish with this crucial example. We use the domain of people:
∃x ∀y isBossOf(x,y)Read this as, ``there is someone who is the boss of everyone.'' From this strong fact we can prove that everyone has a boss, that is, ∀u∃v isBossOf(v,u):
=================================================== ∃x∀y isBossOf(x,y) |− ∀u∃v isBossOf(v,u) 1. ∃x∀y isBossOf(x,y) premise ... 2. b ∀y isBossOf(b,y) assumption ... ... 3. a ... ... 4. isBossOf(b,a) ∀e 2 ... ... 5. ∃v isBossOf(v,a) ∃i 4 ... 6. ∀u∃v isBossOf(v,u) ∀i 3-5 7. ∀u∃v bossOf(v,u) ∃e 1,3-5 ===================================================In the above proof, we let b be our made-up name for the boss-of-everyone. So, we have the assumption that ∀y isBossOf(b,y). Next, we let a be ``anybody at all'' who we might examine in the domain of people. The proof exposes that the boss of ``anybody at all'' in the domain must always be b. ∀i and then ∃i finish the proof.
Here is the proof worked again, with the subproofs swapped:
=================================================== ∃x∀y isBossOf(x,y) |− ∀u∃v isBossOf(v,u) 1. ∃x∀y isBossOf(x,y) premise ... 2. a ... ... 3. b ∀y isBossOf(b,y) assumption ... ... 4. isBossOf(b,a) ∀e 3 ... ... 5. ∃v isBossOf(v,a) ∃i 4 ... 6. ∃v bossOf(v,a) ∃e 1,3-5 7. ∀u∃v isBossOf(v,u) ∀i 2-6 ===================================================
Can we prove the converse? That is, if everyone has a boss, then there is one boss who is the boss of everyone?
∀u∃v isBossOf(v,u) |− ∃x∀y isBossOf(x,y) ???No --- we can try, but we get stuck:
=================================================== 1. ∀u∃v isBossOf(v,u) premise ... 2. a ... 3. ∃v isBossOf(v,a) ∀e 1 ... ... 4. b isBossOf(b,a) assumption 5. ∀y isBoss(b,y) ∀i 2-5 NO --- THIS PROOF IS TRYING TO FINISH THE OUTER SUBPROOF WITHOUT FINISHING THE INNER ONE FIRST. ===================================================We see that the ``block structure'' of the proofs warns us when we are making invalid deductions.
It is interesting that we can prove the following:
∃x∀y isBossOf(x,y) |− ∃z isBossOf(z,z)(``if someone is the boss of everyone, then someone is their own boss'')
=================================================== ∃x∀y isBossOf(x,y) |− ∀u∃v isBossOf(v,u) 1. ∃x∀y isBossOf(x,y) premise ... 2. b ∀y isBossOf(b,y) assumption ... 3. isBossOf(b,b) ∀e 2 ... 4. ∃z isBossOf(z,z) ∃i 4 5. ∃z bossOf(z,z) ∃e 1,2-4 ===================================================Line 3 exposes that the ``big boss,'' b, must be its own boss.
Notice that ∀u∃v isBossOf(v,u) (``everyone has a boss'') holds true for the first two domains but not the third. ∃x∀y isBossOf(x,y) holds true for only the second domain.
Our logic system is designed to work in this way! When we do a logic proof, we are generating new facts that must hold true for any domain for which the premises hold true. This property is called soundness of the logic, and we will examine it more closely in a later section in this chapter.
In this domain, ∀u∃v isBossOf(v,u) holds true as does ∀u∃v isBossOf(u,v) (``everyone bosses someone''), but ∃z isBossOf(z,z) does not hold true.
. . . < -3 < -2 < -1 < 0 < 1 < 2 < 3 < . . .Indeed, one of the main applications of logic is proving properties of numbers. This leads to a famous question: Is it possible to write a collection of premises from which we can deduce (make proofs of) all the logical properties that hold true for the domain of integers?
The answer is NO. In the 1920s, Kurt Goedel, a German PhD student, proved that the integers, along with +, -, *, /, are so complex that it is impossible to ever formulate a finite set (or even an algorithmically defined infinite set) of premises that generate all the true properties of the integers. Goedel's result, known as the First Incompleteness Theorem, set mathematics back on its heels and directly led to the formulation of theoretical computer science (of which this course is one small part). There is more material about Goedel's work at the end of this chapter.
1. Premises premise 2. ∃x P_x premise ... i. a P_a assumption (fill in) ... j. Q (does not mention a!) k. Q ∃e 2,i-j
1. Premises premise . . . i. ...e... (fill in) j. P_e k. ∃x P_x ∃i j
=================================================== board = ... { invariant: isStringArray(board) ∧ len(board) > 0 } def gameOver() : """examines board to see if all squares uncovered. Returns True if so, otherwise returns False.""" { gameOver_pre true } { gameOver_post answer —> ¬(∃ 0 <= i < len(board), board[i] == "covered") ∧ ¬answer —> (∃ 0 <= i < len(board), board[i] == "covered") } answer = True ... while loop that searches board for a board[k] == "covered"; if it finds one, it resets answer = False ... return answer ===================================================The computerized Pat Sajak would use this function like this:
=================================================== done = gameOver() if done : print "We have a winner! Time for a commercial!" else : print "There is still a letter that is covered. Let's continue!" ===================================================Here is the relevant deduction:
=================================================== done = gameOver() { assert: [done/answer]gameOver_post } if done : { 1. done premise 2. [done/answer]gameOver_post premise 3. done —> ¬(∃ 0 < i < len(board), board[i] == "covered") ∧e 2 4. ¬(∃ 0 < i < len(board), board[i] == "covered") —>e 3,1 } print "We have a winner! Time for a commercial!" else : { 1. ¬done premise 2. [done/answer]gameOver_post premise 3. ¬done —> (∃ 0 < i < len(board), board[i] == "covered") ∧e 2 4. ∃ 0 < i < len(board), board[i] == "covered" —>e 3,1 } print "There is still a letter that is covered. Let's continue!" ===================================================Notice that the answer returned by gameOver hides which square on the board is still covered (== "covered").
We repeat an example from a previous chapter to show another use of the existential:
=================================================== def delete(c, s) : """delete locates an occurrence of c in s and removes it and returns the resulting string. If c is not in s, a copy of s is returned, unchanged. """ { pre: isChar(c) ∧ isString(s) } { post: (∃ 0 <= k < len(s), s[k] == c ∧ answer == s[:k] + s[k+1:]) ∨ (∀ 0 <= i < len(s), s[i] != c) ∧ answer == s } index = 0 found = False while index != len(s) and not found : { invariant (∀ 0 <= i < index, s[i] != c) ∧ (found —> s[index] == c) } if s[index] == c : found = True else : index = index + 1 { assert: (index == len(s) ∨ found) ∧ above invariant } if found : answer = s[:index] + s[index+1:] { 1. found premise 2. answer == s[:index] + s[index+1:] premise 3. invariant premise 4. (found —> s[index] == c) ∧e 3 5. s[index] == c —>e 4,1 6. s[index] == c ∧ answer == s[:index] + s[index+1:] ∧i 5,2 7. 0 <= index < len(s) algebra 5 8. ∃ 0 <= k < len(s), s[k] == c ∧ answer == s[:k] + s[k+1:] ∃i 7,6 (where [k/index]) } else : answer = s { 1. ¬found premise 2. answer == s premise 3. (index == len(s) ∨ found) premise 4. invariant premise 5. index == len(s) by P ∨ Q, ¬Q |− P, 3,1 6. ∀ 0 <= i < index, s[i] != c ∧e 4 7. ∀ 0 <= i < len(s), s[i] != c substitution 5,6 8. (∀ 0 <= i < len(s), s[i] != c) ∧ answer == s ∧i 7,2 } return answer ===================================================The ∃i rule is used inside the then-arm of the last conditional, as shown above.
It is important that delete hide the value of its local variable, index, from appearing in its postcondition, because we do not want confusion like this:
=================================================== index = 2 t = "abcd" u = delete("a", t) { at this point, we certainly cannot assert that t[2] = "a"! } ===================================================
{ assert: P } x = e { assert: ∃x_old ( (x == [x_old/x]e) ∧ [x_old/x]P ) }Our use of x_old was hiding the quantifier. Using ∃, we can retain an assertion that uses the old value of x.
In the earlier chapters, we worked examples like this:
=================================================== { x > 0 } x = x + 1 { 1. x_old > 0 premise 2. x == x_old + 1 premise 3. x > 1 algebra 1,2 } ===================================================and noted that x_old must not appear in the last line of the proof.
The above proof is actually the subproof of a proof that finishes with ∃e! Here is the proper proof:
=================================================== { x > 0 } x = x + 1 { 1. ∃x_old(x_old > 0 ∧ x == x_old + 1) premise ... 2. x_old x_old > 0 ∧ x == x_old + 1 assumption ... 3. x_old > 0 ∧e1 2 ... 4. x == x_old + 1 ∧e2 2 ... 5. x > 1 algebra 3,4 6. x > 1 ∃e 1, 2-5 } ===================================================Again, it is crucial that x_old not appear in the assertions on Lines 5 and 6.
We unconsciously use the existential quantifier and ∃e every time we reason about the old, overwritten value of an updated variable.
Also, when we introduce dummy names, like x_in and y_in, to stand for specific values, as an example like this,
=================================================== { assert: x == x_in ∧ y == y_in } temp = x x = y y = temp { assert: x == y_in ∧ y == x_in } ===================================================we are implicitly using existential quantifiers, again, like this:
=================================================== { assert: ∃x_in ∃y_in(x == x_in ∧ y == y_in) } temp = x x = y y = temp { assert: ∃x_in∃y_in(x == y_in ∧ y == x_in) } ===================================================
For example, if we have
{ assert: len(a) > 0 ∧ ∀ 0 <= i < len(a), a[i] > 0 } a[0] = 0we should be able to deduce that
{ assert: a[0] == 0 ∧ len(a) > 0 ∧ ∀ 0 < i < len(a), a[i] > 0 implies: ∀ 0 <= i < len(a), a[i] >= 0 }How can we do this? The existing assignment law is too weak. We require a special law for assignment to array elements. The situation gets delicate if we are unable to deduce the precise numerical value of the index expression, e, in an assignment, a[e] = e'. Unless we can prove otherwise, the assignment has essentially updated ``all'' of a!
Here is an example of a situation where we know nothing about which cell was updated:
=================================================== n = readInt("Type an int between 0 and len(a)-1: ") assert 0 <= n and n < len(a) a[n] = a[n] - 1 ===================================================Clearly only one element of a is decremented. Let's try to reason about this situation:
=================================================== { assert: ∀ 0 <= i < len(a), a[i] > 0 } n = readInt("Type an int between 0 and len(a)-1: ") assert 0 <= n and n < len(a) { assert: 0 <= n ∧ n < len(a) ∧ ∀ 0 <= i < len(a), a[i] > 0 } a[n] = a[n] - 1 { assert: a[n] == a_old[n] - 1 ∧ 0 <= n ∧ n < len(a) ∧ ∀ 0 <= i < len(a), a_old[i] > 0 implies: ??? } ===================================================To move forwards, we must assert that all those variables a[i], such that i != n, retain their old values:
=================================================== a[n] = a[n] - 1 { 1. a[n] == a_old[n] - 1 premise 2. ∀ 0 <= i < len(a), (i!=n) —> a[i] == a_old[i] premise NEW! 3. 0 <= n ∧ n < len(a) premise 4. ∀ 0 <= i < len(a), a_old[i] > 0 premise 5. ... } ===================================================We accept the assertion (premise) on line 2 as a fact, and the forwards assignment law for arrays includes this fact ``for free'' as part of its postcondition.
Here is the law for array assignment:
{ assert: P } a[e] = e' # where e contains _no mention_ of a { 1. a[e] == [a_old/a]e' premise 2. [a_old/a]P premise 3. ∀ 0 <= i < len(a), (i != e) —> a[i] == a_old[i] premise 4. len(a) == len(a_old) premise ... n. Q # must not mention a_old }We gain the new premises in lines 3 and 4. Line 2 is used with ∀e to extract information about array elements that were not affected by the assignment to a[e].
Now we have enough knowledge to make a useful deduction:
=================================================== a[n] = a[n] - 1 { 1. a[n] == a_old[n] - 1 premise 2. ∀ 0 <= i < len(a), (i!=n) —> a[i] == a_old[i] premise 3. 0 <= n ∧ n < len(a) premise 4. ∀ 0 <= i < len(a), a_old[i] > 0 premise 5. a_old[n] > 0 ∀e 4,3 6. a[n] >= 0 algebra 1,5 (next, we salvage the facts about those a[i] such that i != n: ) ... 8. 0 <= x < len(a) assumption ... 9. a_old[x] > 0 ∀e 4,8 ... 10. (x!=n) —> a[x] == a_old[x] ∀e 2,8 ... ... 11. x != n assumption ... ... 12. a[x] == a_old[x] —>e 10,11 ... ... 13. a[x] > 0 algebra 9,12 ... 14. (x!=n) —> a[x] > 0 —>i 11-13 15. ∀ 0 <= x < len(a): (x!=n) —> a[x] > 0 ∀i 8-14 16. a[n] >= 0 ∧ ∀ 0 <= x < len(a): (x!=n) —> a[x] > 0 ∧i 6,15 } ===================================================This tedious proof shows the difficulty in reasoning precisely about an array update with an unknown index value.
It is easy to be discouraged by the length of the above proof, which says that the nth element of a was changed. For this reason, some researchers use a picture notation to encode the assertions. For example, the assertion,
∀ 0 <= i < len(a), a[i] > 0Might be drawn like this:
0 1 ... len(a)-1 +--+--+-- --+--+ a = |>0|>0| ... |>0| +--+--+-- --+--+so that after the assignment, a[n] = a[n]-1, we deduce this new pictorial assertion:
0 1 ... n ... len(a)-1 +--+--+- -+---+ -+--+ a = |>0|>0| ... |>=0| ... |>0| +--+--+- -+---+ -+--+which is meant to portray a[n] >= 0 ∧ ∀ 0 <= x < len(a), (x!=n) —> a[x] > 0.
These pictures can be helpful for informal reasoning, but they quickly get confusing. (For example, where do you draw n's cell in the above picture? What if n == 0? Etc.) Use such drawings with caution.
To summarize, the forwards assignment law for individual array elements reads as follows:
{ assert: P } a[e] = e' # where e contains _no mention_ of a { assert: a[e] == [a_old/a]e' ∧ ∀ 0 <= i < len(a), (i != e) —> a[i] == a_old[i] ∧ len(a) == len(a_old) ∧ [a_old/a]P }
One useful and straightforward technique is selection sort, where the unsorted segment of the array is repeatedly scanned for the smallest element therein, which is extracted at moved to the end of the array's sorted segment.
A trace of a selection sort would look like this:
=================================================== (sorted segment) | (unsorted segment) v a == ["f", "d", "c", "b", "e"] v a == ["b", "d", "c", "f", "e"] ("b" selected and moved to front v by exchanging it with "f") a == ["b", "c", "d", "f", "e",] ("c" selected and moved to front v by exchanging it with "d") a == ["b", "c", "d", "f", "e"] ("d" selected and moved to front by exchanging it with itself) (etc.) | v a == ["b", "c", "d", "e", "f"] (finished) ===================================================We require a function that searches the unsorted segment of the array and locates the position of the least element therein:
=================================================== a = ["e", "d", "a", "c", "b" , "a"] # data structure managed by this module def select(start) : """select returns the index of the smallest element in array a's segment from a[start]...a[len(a)-1].""" { pre: 0 <= start < len(a) } { post: start <= answer < len(a) ∧ ∀ start <= i < len(a), a[answer] <= a[i] } answer = start index = start + 1 { invariant: ∀ start <= i < index, a[answer] <= a[i] } while index != len(a) : if a[index] < a[answer] : answer = index index = index + 1 return answer ===================================================The pre-postconditions tell us the knowledge the function computes.
Next, define these notions of ``ordered'' and ``permuted'' for arrays:
ordered(a) = ∀ 0 < i < len(a), a[i-1] <= a[i] perm(a, b) = (len(a) = len(b)) ∧ (elements of a == elements of b)The second predicate states what it means for one array, a, to have the same elements as another, b, but maybe in a different order. It is a little informal but good enough for us to use here.
The function that does a selection sort uses a loop to repeatedly call select to find the elements to move to the front of the array.
Here's the function and the sketch of the proof. The loop invariant is key --- the elements that have been already selected are moved to the front of a are all guaranteed to be less-than-or-equal-to the elements in a's rear that have not yet been selected:
=================================================== def selectionSort() : """does an in-place sort on global array a, using select.""" { pre true post ordered(a) ∧ perm(a_in, a) (Recall: a_in is the starting value for a) } global a index = 0 { invariant ordered(a[:index]) ∧ perm(a, a_in) ∧ ∀ 0 <= i < index, ∀ index <= j < len(a), a[i] <= a[j] } while index != len(a) : x = select(index) { assert: start <= x < len(a) ∧ ∀ index <= i < len(a), a[x] <= a[i] ∧ invariant } least = a[x] # exchange the least element with the one at the a[x] = a[index] # front of the unsorted segment a[index] = least { assert: ordered(a[:index]) ∧ perm(a, a_in) ∧ index <= x < len(a) ∧ a[index] = least ∧ ∀ index < i < len(a), least <= a[i] implies: ∀ 0 <= i < index, a[i] <= least implies: ordered(a[:index+1]) ∧ ∀ 0 <= i < index+1, ∀ index+1 <= j < len(a): a[i] <= a[j] } index = index + 1 { assert: invariant } ===================================================The key accomplishment of
least = a[x] a[x] = a[index] a[index] = leastis to move the least element in the unsorted suffix of a to the front of that suffix. But that makes the least value eligible to be the rear element of the sorted prefix of a. In this way, the loop's invariant is restored as we finish with index = index + 1.
We can search a sorted array, a, for a value, v, but jumping in the middle of a. If we find v there, we are done. Otherwise, we repeat the step, jumping into the first half or the second half, as needed. Eventually, we find the value (if it is there).
Here is the function, which is famous for its difficulty to write correctly. Glance at it, then read the paragraph underneath it, then return to the function and study its assertions:
=================================================== def search(v, lower, upper) : """searches for value v within array a in the range a[lower]...a[upper]. If found, returns the index where v is; if not found, returns -1""" { search_pre ordered(a) ∧ ∀ 0 <= i < lower, a[i] < v ∧ ∀ upper < j < len(a): v < a[j] (That is, v isn't in a[:lower] and a[upper+1:].) } { search_post ((0 <= answer < len(a)) ∧ a[answer] == v) ∨ (answer = -1 ∧ ∀ 0 <= i < len(a), v != a[i]) } if upper < 0 or lower > len(a)-1 or lower > upper : # empty range to search? { assert: (upper < 0 ∨ lower > len(a)-1 ∨ lower > upper) ∧ search_pre implies: ∀ 0 <= i < len(a), v != a[i] } answer = -1 { assert: answer = -1 ∧ (∀ 0 <= i < len(a), v != a[i]), that is, search_post } else : index = (lower + upper) / 2 if v == a[index] : # found v at a[index] ? answer = index { assert: a[answer] == v implies: search_post } elif v > a[index] : { assert: v > a[index] ∧ search_pre implies: ∀ 0 <= i <= index, a[i] < v implies: [index+1/lower]search_pre } answer = searchFor(v, index+1, upper) { assert: search_post } else : # a[index] < v { assert: a[index] < v ∧ search_pre implies: ∀ index <= j < len(a), v < a[j] implies: [index-1/upper]search_pre } answer = searchFor(v, lower, index-1) { assert: search_post } { assert: search_post } return answer ===================================================To search array, a, for v, we start the function like this:
search(v, 0, len(a)-1)The precondition is the key: to use correctly search(v, lower, upper), we must already know that v is not in a[0]...a[lower-1] and not in a[upper+1}...a[len(a)-1] --- we have already narrowed the search to the range of a[lower]...a[upper]. The function builds on this fact to narrow further the search in subsequent self-calls until v is found or there is an empty range left to search.
The previous two examples display a style of documentation that is used when correctness is critical and one is unable to perform enough testing to generate high confidence that the coding is correctly --- the program must be correct from the first time is it used. Such an approach is taken with safety-critical systems, where money and life depend on the correct functioning of software from the moment it is installed.
To build such a system, we must document the internal structure and connection points of each component so that the system can be connected correctly. This documentation is exactly the pre- and post-conditions for the functions in each component as well as the invariants for the data structures therein.
Here is a small example. It is an implementation of a tic-tac-toe game that follows the usual rules.
First, there is the model module, which models the game board as an array. The game board has an important invariant that ensures that only legal game tokens are placed on the board. There is another data structure in this module that remembers the history of moves made on the board. Both data structures are documented with their invariants. (If you are programming in an object-oriented language and have written a class to model the game board, you call the data-structure invariants, class invariants.
=================================================== """module GameBoard models a Tic-Tac-Toe board. There are two key data structures: --- board, the game board, which holds the players' moves --- history, a list of all the moves made during the game The data structures are managed by calling the functions defined in this module. """ # The game players: X = "X" O = "O" NOBODY = "neither player" ###### The game board, a matrix sized dimension x dimension: EMPTY = "_" # marks an empty square on the board dimension = 3 BOARDSIZE = dimension * dimension # the board itself: board = [] # construct the board with this loop: i = 0 while i != dimension : board.append(dimension * [EMPTY]) i = i + 1 """{ global invariant for board: Only legal markers are placed on it ALL 0 < i,j < dimension, board[i][j] == X v board[i][j] == O v board[i][j] == EMPTY }""" #### A history log of all the moves: it is a list of Marker, Row, Col tuples: history = [] """{ global invariant for history: All moves in history recorded in board forall 0 < i < len(history), history[i]==(m,r,c) and board[r,c] == m }""" ### Functions that manage the board and history: def printBoard() : """prints the board on the display""" """{ pre true post forall 0 <= i,j < dimension, board[i][j] is printed }""" counter = 0 for row in board : for square in row : if square != EMPTY : print square, else : print counter, counter = counter + 1 print print #print history def emptyAt(position) : """examines the ith square on board; returns whether it equals EMPTY. params: position - an int that falls between 0 and the BOARDSIZE returns: whether or not square number position on board is EMPTY """ """{ pre 0 <= position < BOARDSIZE post answer == (board[position/dimension][position%dimension] == EMPTY) }""" answer = False (row,col) = (position/dimension, position%dimension) if 0 <= row and row < dimension and 0 <= col and col < dimension \ and board[row][col] == EMPTY : answer = True return answer def move(marker, position) : """attempts to move marker into the board at position params: marker - a string, should be X or O position -- an int, should be between 0 and the BOARDSIZE """ """{ pre ((marker == X) v (marker == O)) & (0 <= position < BOARDSIZE) post invariants for board and history are maintained }""" global history, board # because we update these global variabes, # we are OBLIGATED to maintain their invariants! if emptyAt(position) : (row,col) = (position/dimension, position%dimension) board[row][col] = marker history = history + [(marker,row,col)] else : pass def winnerIs(mark) : """checks the game board to see if mark is the winner. parameter: mark - a string, should be X or O returns: mark, if it fills a complete row, column, or diagonal of the board; returns NOBODY, otherwise. """ """{ pre (mark == X) v (mark == O) post: (answer == mark --> mark has filled a row or column or diagonal) and (answer == NOBODY) --> mark has not filled any row/col/diag}""" def winnerAlong(vector) : """sees if all the elements in vector are filled by mark""" check = True for index in range(dimension): check = check and (vector[index] == mark) return check # check row i and column i for i in 0,1,...,dimension-1: for i in range(dimension) : columni = [] for j in range(dimension): columni = columni + [board[j][i]] if winnerAlong(board[i]) or winnerAlong(columni) : return mark # check the left and right diagonals: ldiag = [] rdiag = [] for i in range(dimension): ldiag = ldiag + [board[i][i]] rdiag = rdiag + [board[i][(dimension-1)-i]] if winnerAlong(ldiag) or winnerAlong(rdiag) : return mark # else, no winner, so return NOBODY ===================================================The data-structure invariants establish the internal well-formedness of the game board, and the maintenance functions are obligated to preserve and maintain the invariants. In addition, each function is documented with its own pre-post conditions that specify how the function should be called and what the function guarantees if it is called correctly. In the above coding, both informal English and formal logical specifications are written. Whether one writes English or logic depends on how critical absolute correctness might be. (Frankly, some programs, e.g., toys and games, need not be absolutely correct.)
The other module of this little example is the main program --- the controller module --- which enforces the rules of the game, that is, the proper interaction of the game's players with the game board. The controller's main loop has its own invariant that asserts this point. The loop
=================================================== """The Main module controls the tic-tac-toe game.""" import GameBoard from GameBoard import * def readInt(message): """readInt is a helper function that reads an int from the display. If we had a View Module that painted a GUI, this function would be found there. param: message a string returns: an int, denoting the number typed by a player """ """{ pre: message:String post: answer:int }""" needInput = True answer = "" while needInput : text = raw_input(message) if text.isdigit() : answer = int(text) needInput = False return answer player = X # whose turn is it? who goes first? count = 0 # how many moves have been made? winner = NOBODY # who is the winner? """{ loop invariant: The rules of the tic-tac-toe game are enforced: (i) players take turns moving: forall 0 <= i < count, (i % 2)== 0 --> history[i][0] == X and (i % 2)== 1 --> history[i][0] = O (ii) all moves are recorded on board: invariant for history remains true; (iii) board holds only legal game markers: invariant for board remains true (NOTE: (ii) and (iii) should hold _automatically provided that we use the board's maintenance functions.) }""" while winner == NOBODY and count != BOARDSIZE : printBoard() # get the next move: awaitingMove = True while awaitingMove : """{ invariant awaitingMove --> (0 <= m < BOARDSIZE) and emptyAt(m) }""" m = readInt("Player " + player + \ ": type next move (0.." + str(BOARDSIZE) + "): ") if (0 <= m) and (m < BOARDSIZE) and emptyAt(m) : awaitingMove = False # we have received a legal move: """{ assert: ((player == X) v (player == O)) and (0 <= m < BOARDSIZE) and emptyAt(m) implies: [player/marker][m/position]move_pre }""" move(player, m) """{ assert: movepost, that is, invariants for board and history are maintained }""" # determine whether this player is the winner: winner = winnerIs(player) # switch players for the next round: if player == X : player = O else : player = X count = count + 1 """{ assert: loop invariant, all 3 parts, holds }""" # the loop quit, and the game's over: print winner + " won!" printBoard() ===================================================Note how the controller uses the pre-post-conditions for the board's maintenance functions to fulfill its own invariants. In this way, we depend on the documentation from one module to program correctly another. Programming in this style is sometimes called programming by contract.
Logicians have developed a form of predicate logic that omits the existential quantifier and uses instead terms called Skolem functions to name the values represented by each ∃y. A Skolem function is a function name that is used to designate where an existential quantifier should appear. Examples explain the idea best:
=================================================== Everyone has a boss: ∀x ∃y P(x,y) (expressed with a Skolem function, named boss:) ∀x P(x, boss(x)) There is a single boss of everyone: ∃y ∀x P(x,y) (expressed with a Skolem function, named bigb:) ∀x P(x, bigb()) For every husband and wife, there is a minister who married them: ∀x ∀y ∃z M(z,x,y) (expressed with a Skolem function, named minister:) ∀x ∀y M(minister(x,y),x,y) Every two ints can be added into a sum: ∀x ∀y ∃z F(x,y,z) (expressed with a Skolem function, named sum:) ∀x ∀y F(x,y,sum(x,y)) Every boss has a secretary, who talks with everyone: ∀x ∃y (isSec(y) ∧ ∀z talks(x,z)) (expressed with a Skolem function, named s:) ∀x (isSec(s(x)) ∧ ∀z talks(s(x),z)) ===================================================The examples show how the Skolem function acts as a ``witness'' to the missing ∃ without revealing the identity of the individual discussed. You can also see the difference in the first two examples between boss(x) and bigb() --- the first example makes clear how the boss is a function of which x in the domain is considered; the second one makes clear that the boss is independent of all the individuals in the domain.
It is possible to work proofs in predicate logic with Skolem functions. Here are two examples:
=================================================== ∀x isMortal(x) —> hasSoul(x), isMortal(socrates()) |− hasSoul(t()) 1. ∀x isMortal(x) —> hasSoul(x) premise 2. isMortal(socrates()) premise 3. isMortal(socrates()) —> hasSoul(socrates()) ∀e 1 4. hasSoul(socrates()) —>e 3,2 6. hasSoul(t()) def t(): return socrates() ===================================================Here, the individual, Socrates, is represented as a constant Skolem function, socrates(). The key step is the last one, where the desired Skolem function, t(), is defined in terms of socrates():
def t() : return socrates()This function definition takes the place of ∃i.
In the previous example, we could have read isMortal(socrates()) as a shorthand for ∃socrates (isMortal(socrates)). Now, there is no practical difference.
Here is the boss-worker example (``if someone is the boss of everyone, then everyone has a boss''):
=================================================== ∀x isBossOf(x, big()) |− ∀x isBossOf(x, b(x)) 1. ∀x isBossOf(x, big()) premise ... 2. a ... 3. isBossOf(a, big()) ∀e 1 ... 4. hasBoss(a, b(a)) def b(a): return big() 5. ∀x is BossOf(x, b(x)) ∀i 2-4 ===================================================Notice how the defined Skolem function, b(a), disregards its argument and always returns big() as its answer. This is because big() is truly a's boss, no matter what argument is assigned to parameter a.
Like before, it is impossible to prove ∀x isBossOf(x, b(x)) |− ∀x isBossOf(x, big()) --- there is no way to define a Skolem function, def big() : ... b(x) ..., because a value for parameter x is required. In this way, the Skolem functions ``remember'' the placement and use of the original existential quantifiers.
The technical reasoning why Skolem functions work correctly requires so-called Herbrand models and the Henkin completeness theorem. But the programming intuition given here --- defining Skolem functions as Python functions --- works surprisingly well.
A —> B −||− ¬A ∨ B
¬(A ∧ B) −||− ¬A ∨ ¬B ¬(A ∨ B) −||− ¬A ∧ ¬B ¬(∀x A) −||− ∃x ¬A ¬(∃x A) −||− ∀x ¬Aand wherever it appears, replace ¬¬A by A.
Q ∧ (∀x P_x) −||− ∀x (Q ∧ P_x) Q ∨ (∀x P_x) −||− ∀x (Q ∨ P_x)(This is called prenex form.) Now, remove the quantifiers because they are no longer needed (!).
(A ∧ B) ∨ C −||− (A ∨ C) ∧ (B ∨ C)
=================================================== 1. (∃y ∀x isBossOf(y,x)) —> (∃z isBossOf(z,z)) 2. ¬(∃y ∀x isBossOf(y,x)) ∨ (∃z isBossOf(z,z)) 3. (∀y ¬∀x isBossOf(y,x)) ∨ (∃z isBossOf(z,z)) (∀y ∃x ¬isBossOf(y,x)) ∨ (∃z isBossOf(z,z)) 4. (∀y ¬isBossOf(y, x(y)) ∨ isBossOf(z(),z()) 5. ∀y ( ¬isBossOf(y, x(y)) ∨ isBossOf(z(),z()) ) (¬isBossOf(y, x(y))) ∨ isBossOf(z(),z()) 6. (no need to rearrange any ∧s and ∨s) ===================================================At this point, we can apply the resolution algorithm. The resolution rule is slightly modified to handle the variables and Skolem functions: a form of two-way matching, called unification, is used to apply the ∀e rule to variables and Skolem functions:
A ∨ P(E1) ¬P(E2) ∨ B matches = unify(E1,E2) res: --------------------------------------------------------- [matches](A ∨ B)Since P is now a predicate (like isBossOf or isMortal or >), we must unify the arguments E1 and E2 in P(E1) and P(E2) so that they become one and the same. It works like this: For example,
=================================================== for isMortal(soc()) ¬isMortal(z), unify(soc(),z) = [z=soc()] for isMortal(x) ¬isMortal(y), unify(x,y) = [x=y] for isBoss(a(),x) ¬isBoss(y,b()), unify((a(),x), (y,b())) = [y=a(), x=b()] for isBoss(a(),x) ¬isBoss(b(),b()), unify((a(),x), (b(),b())) = FAILURE ===================================================The matches that are computed are applied to the remaining clauses. For example,
¬P(x) ∨ Q(x) P(a()) ------------------------------- Q(a())because matches(x,a()) = [x=a()], so [x=a()]Q(x) is Q(a()).
We apply the revised resolution rule to proving contradictions, like we did with propositional logic. The resolution rule plus unification searches for a witness to a contradiction.
Here are two examples of resolution proofs conducted with unification:
=================================================== ∀x (P(x) —> Q(x)), ∃y P(y) |− ∃z Q(z) The clauses resulting from the two premises and negated goal are ¬P(x) ∨ Q(x), P(y()), ¬Q(z) The resolution proof goes ¬P(x) ∨ Q(x) P(y()) ¬Q(z) | | | +--------------+ | | [x=y()] | Q(y()) | | | +------------------+ | [z=y()] [] ===================================================Here is another example, the workers-and-bosses proof:
=================================================== ∃y ∀x isBossOf(y,x) |− ∀u ∃v isBossOf(v,u) isBossOf(y(),x) ¬isBossOf(v, u()) | | +------------------------+ | [v=y(), x=u()] [] ===================================================Because a clause can be used more than once in a proof, and because a variable can be set to a new value each time it is unified in a resolution step, there is no guarantee that the algorithm will always terminate with success or failure. This is not a flaw of the algorithm --- predicate logic is incomplete in that there can be no algorithm that can decide whether or not a sequent can be proved. Resolution is about as good as we can do in this regard.
At this point, it would be good to review the section on models for propositional logic in Chapter 5. There, we saw that the connectives, ∧, ∨, ¬, —> were understood in terms of truth tables. Also, the primitive propositions were just letters like P, Q, and R, which were interpreted as either True or False.
Within predicate logic, we use predicates, like isMortal() and >, to build propsitions, and we might also use functions, like +, within the predicates. We must give meanings to all predicates and functions so that we can decide whether propositions like isMortal(God) and (3+1)>x are True or False. The act of giving meanings to the predicates and functions is called an interpretation.
Here is an example. Say we have the function symbols, +,-,*,/, and predicate symbols, >,=. What do these names and symbols mean? We must interpret them
These three examples show that the symbols in a logic can be interpreted in multiple different ways. (In Chapter 5, we called an interpretation a ``context.'' In this chapter, we see that a ``context'' is quite complex --- domain, functions, and predicates.)
Here is a second example. There are no functions, and the predicates are isMortal(_), isLeftHanded(_), isMarriedTo(_,_). An interpretation might make all (living) members of the human race as the domain; make isMortal(h) True for every human, h; make isLeftHanded(j) True for exactly those humans, j, who are left handed; and set isMarriedTo(m,f) True for all pairs of humans m, f, who have their marriage document in hand.
You get the idea....
We can ask whether a proposition is True within one specific interpretation, and we can ask whether a proposition is True within all possible interpretations. This leads to the notions of soundness and completeness for predicate logic:
A sequent, P_1, P_2, ..., P_n |− Q is valid in an interpretation, I, provided that when all of P_1, P_2, ..., P_n are True in interpretation I, so is Q. The sequent is valid exactly when it is valid in all possible interpretations. We have these results for the rules of propositional logic plus ∀i, ∀e, ∃i, ∃e:
In the early 20th century, Kurt Gödel showed that it is impossible to formulate a sound set of rules customized for arithmetic that will prove exactly the True facts of arithmetic. Gödel showed this by formulating True propositions in arithmetic notation that talked about the computational power of the proof rules themselves, making it impossible for the proof rules to reason completely about themselves. The form of proposition he coded in logic+arithmetic stated ``I cannot be proved.'' If this proposition is False, it means the proposition can be proved. But this would make the rule set unsound, because it proved a False claim. The only possibility is that the proposition is True (and it cannot be proved). Hence, the proof rules remain sound but are incomplete.
Gödel's construction, called diagonalization, opened the door to the modern theory of computer science, called computability theory, where techniques from logic are used to analyze computer programs. Computability theory tells us what problems computers cannot solve, and why, and so we shouldn't try. (For example, it is impossible to build a program-termination checker that works on all programs --- the checker won't work on itself!) There is also an offshoot of computability theory, called computational complexity theory, that studies what can be solved and how fast an algorithm can solve it.
Given an interpretation of a predicate logic, we can say that the ``meaning'' of a proposition is exactly the set of interpretations (cf. Chapter 5 --- ``contexts'') in which the proposition is True. This returns us to the Boolean-algebra model of logic in Chapter 5. Or, we can organize the interpretations so that an interpretation grows in its domain and knowledge over time. This returns us to the Kripke models of Chapter 5. Or, we can introduce two new programming constructs, the abstract data type and the parametric polymorphic function and extend the Heyting interpretation in Chapter 5.
All of these are possible and are studied in a typical second course on logic.
1. Premises premise ... i. a (fill in) ... j. P_a k. ∀x P_x ∀i i-j
1. Premises premise 2. ∃x P_x premise ... i. a P_a assumption (fill in) ... j. Q (does not mention a!) k. Q ∃e 2,i-j
1. Premises premise 2. ∀x P_x premise . . . j. P_i ∀e 2 (where individual i appears earlier in the proof) (fill in) k. Q
1. Premises premise (fill in) j. P_e k. ∃x P_x ∃i j