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Prove that $ \ \forall x \exists y \ P(x,y) \ $ and $ \ \exists y \forall x \ P(x,y) \ $ are not logically not equivalent in the domain $ \ \{-1,0,1 \} \ $.

Answer:

Let,

$ x=\{-1,0,1 \} \\ y=\{-1,0,1 \} $

Let $ P(x,y) \ $ be the property such that $ x +y \ $ is even number.

Then $ \forall x \exists y \ P(x,y) \ $ is true.

Because,

If $ x=-1 \ $ , then take $ y=1 \ $ such that $ x + y=0 \ $ is even

If $ x=0 \ $ , then take $ y=0 \ $ , such that $ x + y=0 \ $ is even

If $ x=1 \ $ , then take $ y=1 \ $ such that $ x + y =2 \ $ is even

But $ \ \exists y \forall x \ P(x,y) \ $ is $ False $

Am I right ?

I need help.

MAS
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    Correct! Good job! – Bram28 Oct 22 '17 at 12:39
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    Yes this works -- except the notation $x={-1,0,1}$ is wrong and shouldn't be there. And an even easier property to use would be to let $P(x,y)$ be $x=y$. – hmakholm left over Monica Oct 22 '17 at 12:40
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    You could have added why it is true that $\exists y\forall x,P(x,y)$ is false. Otherwise, it's fine. – José Carlos Santos Oct 22 '17 at 12:41
  • can I get any other easier property with ? – MAS Oct 22 '17 at 12:47
  • This is a terrible application of the notation. The operators are commutative in the classical sense. You are instead comparing $\forall x \exists y$ and $\forall x \exists ! y$ which are by definition totally distinct and you dont have to prove anything. – Brethlosze Oct 22 '17 at 12:56
  • @hyprfrcb Everything you said in the previous statement is wrong. – DanielV Oct 22 '17 at 13:15
  • @mabmath I think you mean "for $x \in {-1, 0, 1}$" (similarly for $y$), otherwise that statement doesn't make sense. – DanielV Oct 22 '17 at 13:17
  • mabmath, you were the author of the duplicated question as well. DO NOT EVER REPOST the SAME QUESTION! – amWhy Oct 22 '17 at 16:33
  • Sir it not duplicated , please you can check minutely it that both question differs by the conditions on $ x \ $ and $ \ y \ $. They are different questions. – MAS Oct 22 '17 at 18:06
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    This question is not a duplicate, please unmark it. The other question uses $\exists x \forall y$ instead of $\exists y \forall x$, and it is not just an alpha transform. – DanielV Oct 23 '17 at 02:39
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    Agreed, this is not a duplicate. – Brethlosze Oct 23 '17 at 03:38

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