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Predicate calculus: Proof problem (please check for me) 
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#1
Aug305, 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 nonempty sets. Let D any nonempty set and let p(x,y) be a predicate with a domain of definition D x D. Then since D is nonempty 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 nonempty domain and any predicate p. tehre for (Ax)(∃y)(p(x, y) > p(y, x)) is a tautology. 


#2
Aug305, 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?



#3
Aug405, 01:16 AM

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#4
Aug405, 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 semicounterexample? 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
Aug405, 03:05 AM

P: 5

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)) 


#6
Aug405, 04:46 AM

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[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 nonarbitrary)] 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] 


#7
Aug405, 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) → (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. 


#8
Aug405, 06:15 AM

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I still want to figure out if and how that one rule is valid. 


#9
Aug405, 09:59 AM

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#10
Aug405, 11:17 AM

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


#11
Aug405, 12:38 PM

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