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I am supposed to find a formula for $P \land Q$ using the logical connective $\uparrow$

$P \uparrow Q$ means that not both $P$ and $Q$ is true.

I have already found that
$P \lor Q \equiv (P\uparrow P)\uparrow (Q \uparrow Q)\quad$(1.)
$\neg P \equiv P\uparrow P \quad$ (2.)

I want to use laws like DeMorgan's laws and not use an intuitive argument. Here is my approach:

$P\land Q \equiv \neg (\neg(P\land Q)) \equiv \neg ( \neg P \lor \neg Q)$

From here I can certainly use (1.) and (2.) to come up with a very bulky formula by just applying them over and over, but I can reason myself to that I should be able to get the formula
$(P \uparrow Q)\uparrow (P\uparrow Q)\quad$ (3.)
I just can't figure out how?

PandaDeTapas
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2 Answers2

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$$ (P\uparrow Q)\uparrow(P\uparrow Q)\stackrel{(1)}\equiv \neg((P\uparrow Q)\land(P\uparrow Q))\stackrel{(2)}\equiv \neg(P\uparrow Q)\stackrel{(3)}\equiv \neg(\neg(P\land Q))\stackrel{(4)}\equiv P\land Q$$ where (1) and (3) are the translation of $\uparrow$, (2) is idempotency of $\land$, (4) is double negation.

Hagen von Eitzen
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  • Thank you for your answer. However, I do not understand (3.). What is a "translation"? Or do you just mean that you use common sense since it could be verified with a truth table? – PandaDeTapas Jul 25 '15 at 10:11
  • @PandaDeTapas Tranlation according to "$P\uparrow Q$ means that not both of $P$ and $Q$ are true" – Hagen von Eitzen Jul 25 '15 at 10:55
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By definition, $(P\uparrow Q) = \neg(P \wedge Q)$. Therefore $P \wedge Q = \neg (P\uparrow Q)$. You can now use (2) to get $P \wedge Q = (P\uparrow Q) \uparrow (P\uparrow Q)$.

J.-E. Pin
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