1. ## tautology/logic help

I have a problem that says prove that p^(p exclusive or q) --> (p-->q) is NOT a tautology. I've tried converting both conditional statements into ~p v q but I still get a tautology.

Do I use truth tables to prove that this isn't a tautology? If not by what other means. Or did I just make a mistake somewhere?

2. Originally Posted by guyonfire89
I have a problem that says prove that p^(p exclusive or q) --> (p-->q) is NOT a tautology. I've tried converting both conditional statements into ~p v q but I still get a tautology.
Suppose that $P\text{ is true and }Q\text{ is false }$.
Then $P \wedge \left( {P\underline \vee Q} \right) \text{ is true } ~$ but $~P \to Q \text{ is false }$

3. thanks but I don't understand how P ^ (P xor Q) is true.

4. Originally Posted by guyonfire89
thanks but I don't understand how P ^ (P xor Q) is true.
$P \oplus Q \equiv (P\land\neg Q) \lor(\neg P\land Q)$

5. Originally Posted by guyonfire89
thanks but I don't understand how P ^ (P xor Q) is true.
$\left( {P\underline \vee Q} \right) \equiv \left( {P \wedge \neg Q} \right) \vee \left( {\neg P \wedge Q} \right)$

6. Originally Posted by undefined
$P \oplus Q \equiv (P\land\neg Q) \lor(\neg P\land Q)$
Originally Posted by Plato
$\left( {P\underline \vee Q} \right) \equiv \left( {P \wedge \neg Q} \right) \vee \left( {\neg P \wedge Q} \right)$
Thanks. My teacher never mentioned that in class. Makes a lot more sense