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

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Presentation on theme: "Methods of Proof for Quantifiers Chapter 12 Language, Proof and Logic."— Presentation transcript:

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2 Methods of Proof for Quantifiers Chapter 12 Language, Proof and Logic

3 Valid quantifier steps 12.1 Universal elimination (instantiation): From  xP(x) infer P(c) Existential introduction (generalization): From P(c) infer  xP(x) 1.  x[Cube(x)  Large(x)] 2.  x[Large(x)  LeftOf(x,b)] 3. Cube(d) 4.  x[Large(x)  LeftOf(x,b)] 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. where c is the name of some object of the domain of discourse Let us think about whether there is any similarity with  -elim and  -intro.

4 The method of existential instantiation 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. 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.

5 The method of general conditional proof 12.3.a 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). 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) 1.  xLarge(x) 2.  x[Large(x)  SameRow(x,b)] 3.  xSameRow(x,b) Let us think about whether there is any similarity with  -intro.

6 The method of general conditional proof 12.3.b 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)]. Let us think about why universal generalization in fact makes this rule redundant. 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.

7 Proofs involving mixed quantifiers 12.4.a 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.

8 Proofs involving mixed quantifiers 12.4.b REMEMBER Let P(x), Q(x) be wffs. 1.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.

9 Proofs involving mixed quantifiers 12.4.c 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 { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4194450/slides/slide_9.jpg", "name": "Proofs involving mixed quantifiers 12.4.c Euclid’s Theorem:  x  y[y  x  Prime(y)] Proof.", "description": "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

10 Proofs involving mixed quantifiers 12.4.d 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.


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