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Predicate calculus: Proof problem (please check for me)

by Waylander
Tags: calculus, check, predicate, proof
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Waylander
#1
Aug3-05, 07:27 PM
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Hey guys,
I've been given a problem and I attempted it but I have no idea if it is right, I was hoping you guys could help me.

Show that (∀x)(∃y)(p(x, y) -> p(y, x)) is a tautology.

Since, I am new to logic and Predicate Calculus I was wondering if someone could check my working, (I was amazed how hard it is to get logic 0.0):

Okay, i was told to assume that we will only deal with non-empty sets.

Let D any non-empty set and let p(x,y) be a predicate with a domain of definition D x D. Then since D is non-empty we can find some d ∈ D.
since "p(d,d) -> p(d,d)" is true
We can say there is some y ∈ D such that "p(d, y) -> p(y, d)" is true (namely y=d).
so
(∃y)(p(d, y) -> p(y, d)) is true
and we can also say
(∃y)(p(d, y) -> p(y, d)) is false
(Note: the following is the section I am unsure if I am right or not)
therfore for since
(∃y)(p(d, y) -> p(y, d)) is false
So for any d it must be false.
so
(∃x)(∃y)(p(x, y) -> p(y, x)) must also be false.
(∃x)(∃y)(p(x, y) -> p(y, x)) <=> (Ax)(∃y)(p(x, y) -> p(y, x))
so (∀x)(∃y)(p(x, y) -> p(y, x)) is false too
and we can say
(∀x)(∃y)(p(x, y) -> p(y, x)) must be true
therefore (∀x)(∃y)(p(x, y) -> p(y, x)) must be true for any non-empty domain and any predicate p. tehre for (Ax)(∃y)(p(x, y) -> p(y, x)) is a tautology.
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EnumaElish
#2
Aug3-05, 09:39 PM
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This looks okay to me -- but I don't have a logic degree. Can you take a shortcut and write: "We can say there is some y ∈ D such that "p(d, y) -> p(y, d)" is true (namely y=d) so (∃y)(p(d, y) -> p(y, d)) is true. Since d was arbitraryly chosen, it must be true for all d." That is, can you "merely" rename d as x?
Waylander
#3
Aug4-05, 01:16 AM
P: 5
Quote Quote by EnumaElish
This looks okay to me -- but I don't have a logic degree. Can you take a shortcut and write: "We can say there is some y ∈ D such that "p(d, y) -> p(y, d)" is true (namely y=d) so (∃y)(p(d, y) -> p(y, d)) is true. Since d was arbitraryly chosen, it must be true for all d." That is, can you "merely" rename d as x?
hmm... yeah I guess you could. Thanks. ^_^

honestrosewater
#4
Aug4-05, 01:44 AM
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Predicate calculus: Proof problem (please check for me)

I wish I would have kept going with predicate logic, but what about
[tex]\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ ((x < y) \rightarrow (y < x))[/tex]
?? Even if you mean
[tex](\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ (x < y)) \rightarrow (\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ (y < x))[/tex]
isn't this still a semi-counterexample?

Do you have some inference rules? Normally, to prove that a propostion is a tautology, you derive it from the empty set (of premises). You would use universal and existential generalization, which I'm shaky on, and I'm not sure how to work with predicates either, but it would go something like this:
Start by assuming Pwa, where w is some arbitrary individual; derive Paw; infer, by conditional proof, (Pwa -> Paw); generalize from (Pwa -> Paw) to AxEy(Pxy -> Pyx).
I could be wrong but thought I would mention it just in case.
Waylander
#5
Aug4-05, 03:05 AM
P: 5
Quote Quote by honestrosewater
I wish I would have kept going with predicate logic, but what about
[tex]\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ ((x < y) \rightarrow (y < x))[/tex]
?? Even if you mean
[tex](\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ (x < y)) \rightarrow (\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ (y < x))[/tex]
isn't this still a semi-counterexample?

Do you have some inference rules? Normally, to prove that a propostion is a tautology, you derive it from the empty set (of premises). You would use universal and existential generalization, which I'm shaky on, and I'm not sure how to work with predicates either, but it would go something like this:
Start by assuming Pwa, where w is some arbitrary individual; derive Paw; infer, by conditional proof, (Pwa -> Paw); generalize from (Pwa -> Paw) to AxEy(Pxy -> Pyx).
I could be wrong but thought I would mention it just in case.
Um for the example I think the statement is still true... I think it is saying
For all x: (x < some Natural Number y -> there exists some y < x)
And since this is true for all x then that statement is true for that example, and is not a counter example... I think.

Though those domains and that predicate would be a counter example for:
(∃y)(∀x)(p(x, y) -> p(y, x))
honestrosewater
#6
Aug4-05, 04:46 AM
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Quote Quote by Waylander
Um for the example I think the statement is still true... I think it is saying
For all x: (x < some Natural Number y -> there exists some y < x)
And since this is true for all x then that statement is true for that example, and is not a counter example... I think.
I don't know what that means. Either x and y refer to the same individual, respectively, in both Pxy and Pyx (as in my first statement), or they refer to possibly different individuals (as in my second statement). Of course, this depends on the scope of the quantifiers.

[tex]\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ ((x < y) \rightarrow (y < x))[/tex]

So I can assign x = 1 and y = 2 : (1 < 2) -> (2 < 1). False.

[tex](\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ (x < y)) \rightarrow (\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ (y < x))[/tex]

So I can assign the first x = 3, first y = 4 : 3 < 4. True. And I can assign the second x = 1, but there exists no y in N less than 1, so this and the conditional are false.

You're saying:
1)) Pdd [conditional proof (d is an arbitrary individual)]
2)) Pdd [1, reiteration]
3) Pdd -> Pdd [1, 2 cond. proof]
4) Pdc -> Pcd [3, by what rule? (c is non-arbitrary)]
I don't see how step (4) is valid. How can you substitute c for only some instances of d?
When Pdd is true, Pdc and Pcd will also be true, of course, for at least d = c. However, when Pdd is false, Pdc or Pcd may be true or false, unless you specify that c = d.
Bah, maybe I'm just confusing things, but this really doesn't seem right to me.

Edit: In fact, what's to stop you from doing this:
1) Pdd -> Pdd [premise]
2) Pdc -> Pcd [1, your rule]
3) Pcd -> Pdc [1, your rule?]
4) Pdc <-> Pcd [2, 3, equivalence]
Hurkyl
#7
Aug4-05, 05:47 AM
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Don't forget the rule:

P(a)
----
(∃x):P(x)



Incidentally, in:

[tex]\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ ((x < y) \rightarrow (y < x))[/tex]

If you choose x = 1, I could choose y = 1, and then (x < y) &rarr; (y < x) would be true!

You don't always have to pick y = x, incidentally: for example, P(x, y) could be an equivalence relation, and then you just pick y to be anything at all. But you do need to pick y = x for the general proof, where P is arbitrary.
honestrosewater
#8
Aug4-05, 06:15 AM
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Quote Quote by Hurkyl
Incidentally, in:

[tex]\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ ((x < y) \rightarrow (y < x))[/tex]

If you choose x = 1, I could choose y = 1, and then (x < y) → (y < x) would be true!
Oh, right. Sheesh.

I still want to figure out if and how that one rule is valid.
EnumaElish
#9
Aug4-05, 09:59 AM
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Quote Quote by Hurkyl
If you choose x = 1, I could choose y = 1, and then (x < y) → (y < x) would be true!
Don't you need a weak inequality for this to be true?
honestrosewater
#10
Aug4-05, 11:17 AM
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Quote Quote by EnumaElish
Don't you need a weak inequality for this to be true?
What's a weak inequality - <? I thought Hurkyl was pointing out that (False -> False/True) is true. And since [itex]\forall x (\neg(x < x))[/itex], there exists some y that makes the implication true by making the antecedent false: (y = x).

[tex]\forall x \in \mathbb{N} \ \exists y \in \mathbb{N} \ ((x < y) \rightarrow (y < x))[/tex]
AKG
#11
Aug4-05, 12:38 PM
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  | (no assumptions)
  |----------------
  | 
1 | | p(a,a)                       Assumption
  | |---------
2 | | p(a,a)                       1, Repitition
  |
3 |  p(a,a) --> p(a,a)             1-2, Conditional Introduction
4 | (Ey)(p(a,y) --> p(y,a))        3, Existential Introduction
5 | (Ax)(Ey)(p(x,y) --> p(y,x))    4, Universal Introduction


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