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I was studying for a test and got stuck on this problem for a few days now, I just don't get it.

So given the following sentence for the language $L= < P,R >$, where P and R are unary relation symbols I want to find structures for which the following formula is 1) valid 2) contradictory

$ ((∀x) P(x) -> (∀y) R(y)) -> (∀z) (P(z)->R(z)) $

I believe the answer is very simple, but formulas with unary relations seem to confuse me, any help will be appreciated.

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    What do you mean by 'contradictionary'?2017-01-09
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    Consider two properties that are both true only for some (but not all) the objects of the domain; e.g. domain $\mathbb N$ and $Even(x)$ and $Odd(y)$. In this way, the antecedent is *satisfied* while the consequent is not and thus the formula is not satisfied for the said interpretation.2017-01-09
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    To show that the formula is satisfiable, you have to find an interpretation where the antecedent is *false*: e.g. domain $\mathbb N$ and $Non-negative(x)$ and $Positive(y)$.2017-01-09
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    @MauroALLEGRANZA Thank you a lot Sir, that is exactly what I was looking for, your answer just clarified so many things to me2017-01-09
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    @Bram28 that formula is not valid in structure A2017-01-09
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    @JillWhite A sentence would be valid if it is true in *all* structures, not just one. And it is a contradiction (I think that that is what you may have meant) when it is false in all structures ... Which is not the same is invalid. But to avoid all further confusion, I belive the exercise is simply to find *a* structure in which the snetence is true, and *some* other structure where it is false. Do you still need help with that?2017-01-09
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    @Bram28 Basically what question asks is to find a different structures A,B,C and etc such that $ A⊨ψ $ (ψ is true in B) and $ B⊭ψ $ (or equally ¬ψ is true in A) . The second case is when $ ψ $ is _contradictionary_ in A. But thank you no further help is needed2017-01-09
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    There is no such word as "contradictionary". Google can easily help you spell it.2017-01-10

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