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Discrete Math: Determine whether the statement "for all x, P(x) is true if and only if Q(x) is true" and the statement "for all x, P(x) is true if and only if for all x, Q(x) is true" are logically equivalent. (Intuitively, I'd say they are logically equivalent, but I could always be wrong. Please help, anyone? I could also use help on formally proving an answer)

Mathematics
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The only difference between the two statements is that the second repeats the "for all x" part, but since the first statement implies that we are talking about the same x throughout the expression, that just seems a redundancy, so I agree with you. Can't say I'm positive either though.
Yeah, i asked this question here because a friend told me that they aren't, according to cramster. Of course cramster isn't always right, but it still gave me doubts anyway.
I dont believe they are the same. In the first statement, to check if P(x) is true, you only need to check Q(x). If we were dealing with numbers, this is saying if you wanted to check P(1), you only need to check Q(1) (or vice versa). The second statement is saying something different though. If saying if you want to check if P(x) is true, then check if Q(x) is true for all x. Again with number examples, this is like saying to check if P(1) is true, you have to check Q(1), Q(2), Q(3), etc etc.

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Hm, thank you very much. After thinking about what you said, I came up with a concrete counterexample. Indeed, consider this case: let the domain of x be the set of all integers, let P(x) be the statement that "x>0", and let Q(x) be the statement "x<0." Then the statement "for all x, P(x) is true if and only if Q(x) is true" is clearly false, since an integer can't be greater than or less than 0 at the same time. However, the statement "for all x, P(x) is true if and only if for all x, Q(x) is also true" evaluates to a true statement because both P(x) and Q(x) are false (the iff logical connective requires both logic operands to be both true or both false to make the statement true). Thus, the two statements are not logically equivalent.

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