564 Lecture 8 Sept. 16 1999

1. converse: Relation R is a converse of R'; relations can't just be "converse" by themselves. (It's clear everybody understood this, but in practice, most people listed "converse" as a property of a predicate alone.

Think carefully: is there a converse of "mother of"? Sue is the mother of Bill, Bill is the child of Sue. But: Bill is the child of Dave, is Dave therefore the mother of Bill? Be careful: converse relations must satisfy the biconditional, not just the conditional.

2. Related point: "sister of": some people wrote that "sister of" is "not necessarily" symmetric, others seemed to think that it was "partially" symmetric, or some such thing. While it might be the case that some sisterhood relations are symmetric, all of them certainly are not (as you all noticed, because of brothers), so "sister of" is simply not symmetric, no qualifications necessary. What's confusing is that some relations are not only not symmetric, they are asymmetric, such that no instantiation of them may be symmetric (e.g. taller than). For the purposes of the exercise, it wasn't necessary to distinguish between not symmetric and asymmetric, but it is important to understand that symmetric relations are symmetric universally, for all possible instantiations of their arguments (and more generally, that all those properties must hold universally). "Sister of" would be a symmetric predicate only if the universe consisted entirely of parthenogenetic female persons.

3. Alphabetic variants: again, there were a couple of problems with quantifying the same variable twice but still thinking it means the same thing. E.g.:

Âx(Glitters(x) & ¤x(Gold(x)))

Now, you might think this says, Everything glitters and there is something that glitters and is gold. But no, it says, Everything glitters and there is something that is gold. As soon as you introduce the second quantifier, it's got its own variable, which is subsequently only bound by that one quantifier. Variables may only be bound by one quantifier. A better way of writing the above formula (a clearer alphabetic variant) would be:

Âx(Glitters(x) & ¤y(Gold(y)))

Keep in mind when you're writing your formulas that a quantifier can and must bind any appropriate variable in its scope, and that any given variable can only be bound by one quantifier.

We have our rules of inference from the propositional logic, and they still apply as well as ever. The problem is to get the quantifiers out of formulas and replace variables with constants, so that we have atomic propositions that we can apply our rules of inference to. Now, we only want to do this in such a way that we preserve the truth conditions of the whole, quantified proposition, so we have to be very careful. To help you avoid making mistakes, deSwart only gives two rules of inference for stripping away quantifiers, but in fact there are four, all of which you will need to do the homework.

4. Universal Instantiation:

Âx(P(x))

P(c)

where c is an arbitrarily chosen constant.

This is a rule which replaces a variable with a constant in a universally quantified formula. This works out all right because a universally quantified formula is true if and only if every possible instantiation of an object from U makes the formula true. So we're safe if we replace a universally quantified variable with an arbitrary constant; it's guaranteed to preserve the truth value of the formula.

This is how we can, at last, prove that if Socrates is a man, then he is mortal, from the premise, All men are mortal.

Here's the argument in natural language and predicate calculus:

5.

All men are mortal. Âx(Man(x)->Mortal(x))

Socrates is a man. Man(s)

Socrates is mortal. Mortal(s)

And here's the proof in predicate calculus:

6. 1) Âx(Man(x)->Mortal(x)) Premise

2) Man(s) Premise

3) Man(s)->Mortal(s) 1, Universal Instantiation

4) Mortal(s) 2,3, Modus Ponens

Now, consider the simple fact of universally instantiating line (3), without worrying about the rest of the proof. Because it was a universally quantified formula, it was safe to chose a constant to instantiate it. It could have been any constant at all. For the purposes of this proof, we chose Socrates, because we had another formula that already included Socrates. However, we could have chosen any constant whatever. That is, we could have chosen an arbitrarily selected constant. Now, anything that is true of an arbitrarily selected constant (which we'll call c) is true of everything in the universe of discourse. So, if you've gone from universal quantification to an arbitrarily selected constant, it's safe to go back to universal quantification. This procedure is called Universal Generalization:

7. Universal Generalization:

P(c)

Âx(P(x))

You need to use this inference in arguments of the following form:

8.

Every rabbit is a quadruped Âx(Rabbit(x)->Quadruped(x))

Every quadruped is warm-blooded Âx(Quadruped(x)->Warm-b(x))

Every rabbit is warm-blooded Âx(Rabbit(x)->Warm-b(x))

And here's the proof in predicate calculus:

9. 1) Âx(R(x)->Q(x)) Premise

2) Âx(Q(x)->W(x) Premise

3) R(c)->Q(c) 1, Universal Instantiation

4) Q(c)->W(c) 2, Universal Instantiation

5) R(c)->W(c) 3,4 Hypothetical Syllogism

6) Âx(R(x)->W(x)) 5, Universal Generalization.

There are two key elements in this proof. The first is that we chose the same arbitrary constant to instantiate the variable in 3 and 4. Again, this is legitimate because both formulas are universally quantified and are true for any constant in the discourse, so they'll be true for the same constant. This enables us to use the regular rule of inference, Hypothetical Syllogism in line 5) to prove that any particular rabbit is warm blooded. Then, since the constant c, wherever it was introduced, was in all cases was an arbitrary constant, we are justified in generalizing back to the universal case in line 6. Basically, you can use U.I. to remove a quantifier, and U.G. to replace a quantifier afterward provided the constant that is being replaced is arbitrary in all instances throughout the proof. Consider the proof about Socrates' mortality above. What would go wrong if we replaced the constant s, introduced by Universal Instantiation in line 3), with a variable in the concluding line, giving Âx(Mortal(x), and stating that all things are mortal? This is not a valid conclusion from the premises that "All men are mortal" and "Socrates is a man". What went wrong?

Now, the next completely safe rule of inference:

10. Existential Generalization

P(c)

¤x(P(x))

This simply states that if some constant is a member of the set denoted by some predicate, then there is something that instantiates that predicate. (Remember "Jane brought a cake" entails "Someone brought a cake"?) Here's a natural language argument, followed by its proof, using E.G.

11.

Jill is human. Human(j)

All humans are mortal Âx(Human(x)->Mortal(x))

There is at least one mortal. ¤x(Mortal(x))

12. 1) H(j) given

2) Âx(H(x)->M(x)) given

3) H(j)->M(j) 2, U.I.

4) M(j) 1,3, M.P.

5) ¤x(M(x)) 4, E.G.

Now, the trickiest rule of inference to use correctly, Existential Instantiation. It looks like this:

13. Existential Instantiation

¤x(P(x)

P(d)

What it does is strip away the existential quantifier by replacing the variable with one of the constants that instantiates the predicate. There might only be one. What the existential quantifier does is assert that there is at least one individual in the universe of discourse for whom the predicate is true. What Existential Instantiation does is replace the variable with that constant. So this constant is not arbitrary; it's from the subset of individuals that instantiate the predicate.

What makes it especially tricky is if you're going to apply it twice. Let's say that you have premises of the following form:

14. How Not To Apply E. I.

Some cats are vicious. ¤x(Cat(x) & Vicious(x))

Some dogs are vicious. ¤x(Dog(x) & Vicious(x))

Some cats are dogs. ¤x(Cat(x) & Dog(x)

15. The Invalid Proof of 14:

1) ¤x(Cat(x) & Vicious(x)) Premise

2) ¤x(Dog(x) & Vicious(x)) Premise

3) Cat(d) & Vicious(d) 1, E.I.

4) Dog(d) & Vicious(d) 2, E.I. ¡Incorrect!

5) Cat(d) 3, Simplification.

6) Dog(d) 4, Simplification

7) Cat(d)&Ddog(d) 5,6 Conjunction

8) ¤x(C(x) & D(x))

What went wrong? Essentially, in line 4, when we chose to instantiate the set of dogs with the constant (d), right then we were asserting there was a particular cat d who was also the dog d. There is no guarantee that the sets denoted by two existentially quantified predicates will overlap in that way, so choosing the same constant to instantiate both predicates was a horrible mistake. So if you're going to chose a constant to instantiate an existential quantification, i.e. if you're going to apply Existential Instantiation, you must use a constant that has not occurred in any previous line of the proof!!

16. The Condition on Existential Instantiation:

Here's a good argument whose proof uses Existential Instantiation correctly, in clever combination with Universal Instantiation, (U.I has no such restriction, of course; remember that U.G. does, though, if a constant isn't selected arbitrarily):

17. Some toadstools are poisonous. ¤x(T(x)&P(x))

All poisonous things are harmful. Âx(P(x)->H(x))

Some toadstools are harmful. ¤x(T(x)&H(x))

18. 1) ¤x(T(x)&P(x)) Given

2) Âx(P(x)->H(x)) Given

3) T(d) & P(d) 1, E.I.

4) P(d)->H(d) 2, U.I. <–This is ok because U.I. unrestricted!

5) P(d) 3, Simp.

6) H(d) 5,4, M.P

7) T(d) 3, Simp.

8) T(d)&H(d) 6,7, Conjunction

9) ¤x(T(x)&H(x)) 8, Existential Generalization.

Note! It would not be legitimate to use Universal Generalization on line 8), even though we introduced d in line 3 by U.I., because it wasn't arbitrary – d had occurred in a non-arbitrary usage on a previous line of the proof, just like s had in the proof about Socrates above.

A further caution when using these rules. You can only introduce quantifiers by means of these rules that have the entire formula in their scope. That is why it's important to be able to derive PNFs for formulas. Before taking out or putting in any variables and their quantifiers, you must be sure that the particular quantifier takes scope over the whole formula. This is particularly important when the formula has a conditional in it; remember how tricky the interaction of conditionals with quantifiers is! Here's an example where if you don't put the quantifier at the very beginning, something goes wrong:

19. (The natural language versions of premise is: "If Rush is a radio host, then everyone is qualified")

1) Âx(Host(r)->Qualified(x)) Premise

2) H(r)->Q(c) 1, U.I.

3) | H(r) Auxiliary Premise (we're proving a conditional)

4) | Q(c) 2,3, M.P.

5) |Âx(Q(x)) 4, U.G. (legitimate!)

6) H(r)->Âx(Q(x)) 3-5, conditional proof

7) ¤x(H(x)->Âx(Q(x))) 6, Existential Generalization.

Wrong 7') ¤x(P(x))->Âx(Q(x))

Problem with 7'): Existential quantifier only has scope over antecedent!!

7) says "There is someone such that, if he is a radio host, then everyone is qualified" (which is valid from our premise). 7') says, "If there is a radio host, then everyone is qualified" (which is not valid from our premise). The difference comes because in 7') the existential quantifier only has scope over the antecedent.

So, before applying any instantiation or generalization to a formula, be sure any relevant quantifiers are at the front of the formula, which you'll have to use the equivalences given earlier to accomplish, massaging the formula until the quantifier pops out the front.

The final cautionary tale with respect to instantiation and generalization: it is important to ensure, when you're instantiating both existentially quantified and universally quantified elements in a formula, that you "undo" the instatiating with generalizing in the opposite order. That is, whatever instantiation you did most recently, you must generalize that constant first. Otherwise, you'll get the quantifiers out of order (because whenever you generalize, you put the quantifier at the beginning of the formula, to give it scope over the whole thing).

Here's an argument where something can go wrong if you generalize back from instantiating in the wrong order:

20. (F(x,y) = x is father of y)

Every human has a father Âx¤y(H(x)->F(y,x))

All Bulgarians are humans Âx(B(x)->H(x))

Every Bulgarian has a father Âx¤y(B(x)->F(y,x))

21. 1) Âx¤y(H(x)->F(y,x))

2) Âx(B(x)->H(x))

3) ¤y(H(c)->F(y,c)) 1, U.I.

4) H(c)->F(d,c) 3, E.I.

5) B(c)->H(c) 2, U.I.

6) B(c)->F(d,c) 5,4, H.S.

7) ¤x(B(c)->F(x,c)) 6, E.G.

8) Ây¤x(B(y)->F(x,y)) 7, U.G.

This says, "for all y, there is an x such that if y is Bulgarian, then x is the father of y", or in natural English, "Every Bulgarian has a father"

Now, consider what would happen if we did the existential generalization after the universal generalization, replace lines 7) and 8) as follows:

7') Âx(B(x)->F(d,x)) U.G.

8') ¤yÂx(B(x)->F(y,x)) E.G.

This actually says, "there is a y such that for all x, if x is bulgarian, y is the father of x" or, more colloquially "There is someone who is the father of all Bulgarians". Oops!

6.1 (i) G(x₁, x₂, x₃)

1 [| G(x₁, x₂, x₃) |]^M,g2 = 1 iff < [|x₁|]^M,g2, [|x₂|]^M,g2, [|x₃|]^M,g2> E [|G|]^M,g2

2 [|G|]^M,g2 = {<x,y,z> E UxUxU | x+y=z }

3 [|x₁|]^M,g2 = g₂(x₁) = 1

4 [|x₂|]^M,g2 = g₂(x₂) = 5

5 [|x₃|]^M,g2 = g₂(x₃) = 6

6 < [|x₁|]^M,g2, [|x₂|]^M,g2, [|x₃|]^M,g2> = <1,5,6> 3,4,5

7 <1,5,6> E {<x,y,z> E UxUxU | x+y=z } 6

8 < [|x₁|]^M,g2, [|x₂|]^M,g2, [|x₃|]^M,g2> E [|G|]^M,g2 6, 2

9 [| G(x₁, x₂, x₃) |]^M,g2 = 1 8, 1, M.P.

(ii) Âx₁G(x₁, j, x₁)

1. [| Âx₁G(x₁, j, x₁) |]^M,g2 = 1 iff [| G(x₁, j, x₁) |]^M,g{x1/e] = 1 for all individuals e in the discourse:

1'. ... iff [| G(x₁, j, x₁) |]^M,g{x1/e] =1 when g^x1/e(x₁) = 0 and [| G(x₁, j, x₁) |]^M,g{x1/e] =1when g^x1/e(x₁) = 1, and [| G(x₁, j, x₁) |]^M,g{x1/e] =1 when g^x1/e(x₁) = 2, and [| G(x₁, j, x₁) |]^M,g{x1/e] =1 when g^x1/e(x₁) = 3, and [| G(x₁, j, x₁) |]^M,g{x1/e] =1 when g^x1/e(x₁)=4 and [| G(x₁, j, x₁) |]^M,g{x1/e] =1 when g^x1/e(x₁) =5 and [| G(x₁, j, x₁) |]^M,g{x1/e] =1 when g^x1/e(x₁) =6 and [| G(x₁, j, x₁) |]^M,g{x1/e] =1 when g^x1/e(x₁) =7 and [| G(x₁, j, x₁)|]^M,g{x1/e] =1 when g^x1/e(x₁) = 8 and [| G(x₁, j, x₁) |]^M,g{x1/e] =1 when g^x1/e(x₁) = 9.

3. [|j|] = 0

4. 0+0=0 and 1+0=1 and 2+0=2 and 3+0=3 and 4+0=4 and 5+0= 5 and 6+0=6 and 7+0=7 and 8+0=8 and 9+0=9.

5. <0,0,0> E {<x,y,z> E UxUxU | x+y=z } and <1,0,1> E {<x,y,z> E UxUxU | x+y=z } and <2,0,2> E {<x,y,z> E UxUxU | x+y=z } and <3,0,3> E {<x,y,z> E UxUxU | x+y=z }and <4,0,4> E {<x,y,z> E UxUxU | x+y=z } and <5,0,5> E {<x,y,z> E UxUxU | x+y=z } and <6,0,6> E {<x,y,z> E UxUxU | x+y=z } and <7,0,7> E {<x,y,z> E UxUxU | x+y=z }and <8,0,8> E {<x,y,z> E UxUxU | x+y=z }and <9,0,9> E {<x,y,z> E UxUxU | x+y=z }

6. [| Âx₁G(x₁, j, x₁) |]^M,g2 = 1

(ii) Âx₂¤x₃ [G(x₁, x₂, x₃)]

1. [| Âx₂¤x₃ [G(x₁, x₂, x₃)] |]^M,g2 = 1 iff [| ¤x₃ [G(x₁, x₂, x₃) |]^M,g{x2/e] =1 for all individuals e in the universe of discourse.

2. For g^x2/e(x₂)=0, [| ¤x₃ [G(x₁, x₂, x₃) |]^M,g{x2/0] =1 iff [| [G(x₁, x₂, x₃) |]^M,g[x3/f] =1 for at least one individual f in the universe of discourse.

3. [|x₁|]^M,g2 = g₂(x₁) = 1

4. [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 1 iff <1,0, [|x₁|]^M,g[x3/f] > E {<x,y,z> E UxUxU | x+y=z }

5. Let g^x3/f(x₃) = 1. [| [G(x₁, x₂, x₃) |]^M,g[x3/f] =1.

(because <1,0,1>E {<x,y,z> E UxUxU | x+y=z }

6. For g^x2/e(x₂)=0, [| ¤x₃ [G(x₁, x₂, x₃) |]^M,g{x2/0] =1.

7. For g^x2/e(x₂)=9, [| ¤x₃ [G(x₁, x₂, x₃) |]^M,g{x2/0] =1 iff [| [G(x₁, x₂, x₃) |]^M,g[x3/f] =1 for at least one individual f in the universe of discourse.

8. [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 1 iff <1,9, [|x₁|]^M,g[x3/f] > E {<x,y,z> E UxUxU | x+y=z }

9. Let g^x3/f(x₃)=0. [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,0> ~E {<x,y,z> E UxUxU | x+y=z }

10. Let g^x3/f(x₃)=1. [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,1> ~E {<x,y,z> E UxUxU | x+y=z }

11. Letg^x3/f(x₃)=2. [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,2> ~E {<x,y,z> E UxUxU | x+y=z }

12. Let g^x3/f(x₃)=3. [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,3> ~E {<x,y,z> E UxUxU | x+y=z }

13. Let g^x3/f(x₃)=4 [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,4> ~E {<x,y,z> E UxUxU | x+y=z }

14. Let g^x3/f(x₃)=5 [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,5> ~E {<x,y,z> E UxUxU | x+y=z }

15. Let g^x3/f(x₃)=6 [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,6> ~E {<x,y,z> E UxUxU | x+y=z }

16. Let g^x3/f(x₃)=7 [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,7> ~E {<x,y,z> E UxUxU | x+y=z }

17. Let g^x3/f(x₃)=8 [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,8> ~E {<x,y,z> E UxUxU | x+y=z }

18. Let g^x3/f(x₃)=9 [| [G(x₁, x₂, x₃) |]^M,g[x3/f] = 0.

(because <1,9,9> ~E {<x,y,z> E UxUxU | x+y=z }

19. For g^x2/e(x₂)=9, [| ¤x₃ [G(x₁, x₂, x₃) |]^M,g{x2/0] =0

20. [| Âx₂¤x₃ [G(x₁, x₂, x₃)] |]^M,g2 = 0

(iv) Âx₁¤x₂(K(x₁,x₂) & Ø x₁=x₂)

For all x₁ there is an x₂ such that x₁¾x₂ and x₁ is not equal to x₂

This is false in M₂ w/r to g₂ because for x₁=9 there is no x₂ such that x₂ is not 9 and yet x₁, i.e. 9, is less than or equal to x₂.

(v) ¤x₁(Q(x₁) & Ø P(x₁)) & ¤x₁(Q(x₁) & P(x₁))

There is a number which is a number Bill Clinton uses in his lottery forms and is not odd, and there is another number which is a number Bill Clinton uses in his lottery forms and is odd.

This is true, because 8 is a number Bill Clinton uses in his lottery forms and it is not an odd number, and 3 is a number Bill Clinton uses in his lottery forms and it is an odd number.

(vi) Âx₁((P(x₁) & x₁=m) -> Q(x₁))

For all numbers, if a number is odd and not equal to 9, then it is a number that Bill Clinton uses on his lottery form.

This is true, because 9 is the only odd number BC doesn't use on his lottery form.

(vii) Âx₁(Q(x₁) -> (P(x₁) & Ø x₁=m))

For all numbers, if a number is used by BC on his lottery form, then the number is odd and it is not equal to 9.

This is false, because BC uses even numbers on his lottery form too.

....

(xi) Âx₁(P(x₁) Ú Ø P(x₁))

(xii) Âx₁(G(x₁,x₁,x₁) -> x₁=j)

....

(xv) Âx₁¤x₂¤x₃¤x₄¤x₅((Q(x₁) & P(x₁))->(x₁=x₂ Ú x₁=x₃ Ú x₁=x₄ Ú x₁=x₅))