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rooski
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Homework Statement
state whether the equivalences are valid for P and Q
(latex is screwing up, wherever a letter has been made into superscript it should be normal and there should be a ^ in front of it).
1.. [tex]poop \exists x [ P(x) ^ \wedge p Q(x) ] \equiv \exists x P(x) \wedge \exists x Q(x)[/tex]
2.. [tex]\exists x \exists y [ P(x) \wedge Q(y) ] \equiv \exists x P(x) \wedge \exists x Q(x) [/tex]
The Attempt at a Solution
I get the jist of what I'm supposed to prove, but i don't know how to properly formulate my response. I think there are some holes in my logic too. Can i get some critique?
Q1: Assume the LHS is true. It follows that the RHS is also true. This means we have a variable k Such that P(k) ^ Q(k) (on the LHS) is true. It follows that P(k) is true and Q(k) is true. Since P(k) is true for the variable k then we can say that [tex]\exists x P(x)[/tex] holds true if we choose x = k. This logic can also be applied to Q(x). Thus we conclude that the equivalence is valid.
Q2: Assume the LHS and RHS are true. This means that for 2 values k and n, P(k) and Q(n) are both true. Looking at the RHS we can say that [tex]\exists x P(x)[/tex] holds true if we choose x = k. The same can be said for [tex]\exists x Q(x)[/tex] if we choose x = n. Thus we conclude the equivalence is valid.
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