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I have the following: p(X) → NOT p(X)

The exercise requires to find an interpretation that makes the expression true and one that makes the expression false.

So my reasoning was this, obviously if the rhs is true the lhs will be false and viceversa, so keeping in mind the truth table for logical implication I would want the left side to be false and the right to be true for the whole expression to be true (right?)

So for my first interpretation I take {a} as domain. And set p(a) as false, which makes the right side true and the expression true.

For the second interpretation I still take {a} as domain but set p(a) to be true, which would make the right side false and the expression false.

However the solutions say the opposite. As in p(a) being true makes the expression true and p(a) being false makes the expression false.

Can anyone help me with this? I don't get what I'm missing...

user154562
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1 Answers1

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You're correct, and the solution manual is wrong. Note that $$p(a) \rightarrow \lnot p(a) \equiv \lnot p(a) \lor \lnot p(a) \equiv \lnot p(a)$$

...which is true if and only if $p(a)$ is false.

amWhy
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  • That's great, thanks! Been wasting so much time on that one. – user154562 Jun 01 '14 at 13:18
  • This answer is correct. That said, the first equivalence here only applies for two-valued logic. The OP's answer will work for both two-valued logic, Lukasieiwicz three-valued and n-valued logic, and Lukasiewicz infinite-valued logic. – Doug Spoonwood Jun 01 '14 at 13:24