Methods of Proof for Quantifiers

1. Language, Proof and Logic Methods of Proof for Quantifiers Chapter 12

2. 12.1 Valid quantifier steps Universal elimination (instantiation): From xP(x) infer P(c) where c is the name of some object of the domain of discourse Existential introduction (generalization): From P(c) infer xP(x) 3 says that d is a cube. And 1 says that all cubes are large. Thus, d is large. But 2 says that every large object is to the left of b. So, d is to the left of b. To summarize, d is large and is to the left of b. Thus, there is a large object to the left of b. 1. x[Cube(x)Large(x)] 2. x[Large(x)LeftOf(x,b)] 3. Cube(d) 4. x[Large(x)LeftOf(x,b)] Let us think about whether there is any similarity with -elim and-intro.

3. 12.2 Existential instantiation (elimination): Once you have proven xP(x) (or have it as a premise), you can select a “neutral” (not used elsewhere) name d and use P(d) as a valid assumption. The method of existential instantiation 1. x[Cube(x)Large(x)] 2. x[Large(x)LeftOf(x,b)] 3. xCube(x) 4. x[Large(x)LeftOf(x,b)] 3 says that there is a cube. Let d be such a cube, i.e. assume Cube(d) (is true). 1 says that all cubes are large. Thus, d is large. But 2 says that every large object is to the left of b. So, d is to the left of b. To summarize, d is large and is to the left of b. Thus, there is a large object to the left of b. Important: If we had selected d=b, we would have been able to “prove” xLeftOf(x,x)! Let us think about whether there is any similarity with-elim.

4. 12.3.a The method of general conditional proof Universal generalization (introduction): Once you have proven P(d) for some “neutral” (not used elsewhere) name d (denoting a “totally arbitrary” object), you can conclude xP(x). • 1. xLarge(x) • 2. x[Large(x)SameRow(x,b)] • 3. xSameRow(x,b) Consider any object d. By 1, d is large. But, by 2, every large object is in the same row as b. So, d is in the same row as b. As d was arbitrary, we conclude that every object is in the same row as b. Important: The “arbitrary” object 1. Cube(b) d indeed has to be arbitrary. Things 2. x[Cube(x)Large(x)] will go wrong if you select d=b here 3. xLarge(x) Let us think about whether there is any similarity with-intro.

5. 12.3.b The method of general conditional proof General conditional proof: Once you have proven Q(d) from the assumption P(d) for some “neutral” (not used elsewhere) name d (denoting a “totally arbitrary” object), you can conclude x[P(x)Q(x)]. • 1. x[Cube(x)SameRow(x,b)] • 2. x[SameRow(x,b)Small(x)] • 3. x[Cube(x)Small(x)] Consider any object d, and assume d is a cube. 1 says that every cube is in the same row as b. So, d is in the same row as b. But, by 2, everything in the same row as b is small. So, d is small. As d was arbitrary, we conclude that every cube is small. Let us think about why universal generalization in fact makes this rule redundant.

6. 12.4.a Proofs involving mixed quantifiers 1.y[Girl(y)  x(Boy(x)  Likes(x,y))] 2.x[Boy(x) y(Girl(y)  Likes(x,y))] Consider an arbitrary boy d. By 1, there is a girl who is liked by every boy. Let c be such a girl. So, d likes c. That is, d likes some girl. As d was arbitrary, we conclude that every boy likes some girl. 1.x[Boy(x) y(Girl(y)  Likes(x,y))] 2. y[Girl(y)  x(Boy(x)  Likes(x,y))] Pseudo-proof: Consider an arbitrary boy d. By 1, d likes some girl. Let c be such a girl. Thus, d likes c. Since d was arbitrary, we conclude that every boy likes c. So, there is a girl (specifically, c) who is liked by every boy.

7. 12.4.b Proofs involving mixed quantifiers • REMEMBER • Let P(x), Q(x) be wffs. • Existential Instantiation: If you have proven xP(x) then you may • choose a new constant symbol c to stand for any object satisfying • P(x) and so you may assume P(c). • 2. General Conditional Proof: If you want to prove x[P(x)Q(x)] • then you may choose a new constant symbol c, assume P(c), and • prove Q(c), making sure that Q does not contain any names • introduced by existential instantiation after the assumption of P(c). • 3. Universal Generalization: If you want to prove xQ(x) then you • may choose a new constant symbol c and prove Q(c), making sure • that Q does not contain any names introduced by existential • instantiation after the introduction of c.

8. 12.4.c Proofs involving mixed quantifiers Euclid’s Theorem: xy[yx  Prime(y)] Proof. Consider an arbitrary natural number n. Our goal is to show that y[yn  Prime(y)],from which Euclid’s theorem follows by universal generalization. Let k be the product of all the prime numbers less than n. Thus each prime with <n divides k without remainder. Now let m=k+1. Each prime less than n divides m with remainder 1. But we know that m can be factored into primes. Let p be one of those primes. Clearly, by the earlier observation, pn. Hence, by existential generalization, there is a prime (specifically, p) greater or equal to n. As n was arbitrary, we conclude that xy[yx  Prime(y)].

9. 12.4.d Proofs involving mixed quantifiers The Barber Paradox: xy [Shave(x,y) Shave(y,y)] The domain of discourse is the set of all men in a small village. Proof. Assume, for a contradiction, that 1. xy [Shave(x,y)  Shave(y,y)] Let b be a man (barber) such that 2.y [Shave(b,y)  Shave(y,y)] is true. By universal instantiation from 2, 3. Shave(b,b)  Shave(b,b). But this is (indeed) a contradiction.