Can anyone help my to understand how to find a model for show that this is not an Logical Equivalence?
i have no idea of which kind of dominion i've to use.
really thanks to everyone from spain
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Let the domain be $\{a,b\}$. Suppose $P(a)$ is true, $P(b)$ is false, and $Q(a)$, $Q(b)$ are both false.
Then $\exists x(P(x)\rightarrow Q(x))$ is true. Pick $x=b$.
But it is clear that $\exists xP(x)\rightarrow \exists x Q(x)$ is false.
André Nicolas
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Thank you Andrè Nicolas, so p(a) is a property true for a, and p(b) is a property false for b. And, q(a) and q(b) are two properties false for a and b, right? don't i need to specify what are this properties? – Bastian Muller Feb 24 '14 at 18:20
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Well, we are in the predicate calculus, so $P$ is a single property, which holds or fails to hold at various elements of the domain. For example (but not for your problem) the domain might be the positive integers and $P(x)$ the property "$x$ is prime." So $P(17)$ is true and $P(18)$ is false. – André Nicolas Feb 24 '14 at 18:26
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ok so we are in "our" dominion , so because we have only {a,b} we can use p(a) or p(b) for p(x) and q(a) or q(b) for q(x), am i right Andrè? – Bastian Muller Feb 24 '14 at 18:32
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Yes, the domain (set that variables range over) is the set that only contains the objects $a$ and $b$, – André Nicolas Feb 24 '14 at 18:34
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perfect, thank you again :) – Bastian Muller Feb 24 '14 at 18:59
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You are welcome. The answer used a "peculiarity" of $p\rightarrow q$: It is true whenever $p$ is false. – André Nicolas Feb 24 '14 at 19:05
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He answered well, but also do note that just because x is in both sides, it doesn't mean it's the same variable. For example, since the left side has parenthesis, it means there is a x that is assigned to both P and Q. But on the right side, it means there is an x that is in P, but it also means there is an x in Q. It doesn't mean both x in P and Q are the same x being used. – Hedylove Oct 07 '15 at 14:34