Predicate calculus: Proof problem (please check for me)

  • Thread starter Waylander
  • Start date
5
0

Main Question or Discussion Point

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.
 
Last edited:

Answers and Replies

EnumaElish
Science Advisor
Homework Helper
2,285
123
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?
 
5
0
EnumaElish said:
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
Gold Member
2,071
5
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.
 
5
0
honestrosewater said:
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
Gold Member
2,071
5
Waylander said:
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]
 
Last edited:
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,843
17
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
Gold Member
2,071
5
Hurkyl said:
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. :redface:

I still want to figure out if and how that one rule is valid.
 
EnumaElish
Science Advisor
Homework Helper
2,285
123
Hurkyl said:
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
Gold Member
2,071
5
EnumaElish said:
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
Science Advisor
Homework Helper
2,559
3
Code:
  | (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
 

Related Threads for: Predicate calculus: Proof problem (please check for me)

  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
7
Views
2K
Replies
1
Views
523
  • Last Post
Replies
4
Views
2K
Top