# Thread: Showing equality via a simple calculation

1. ## Showing equality via a simple calculation

I'm studying for an introductory course on logic, I'm at the point where I've to show the equality of certain formulas using a calculation with some basic standard equivalences. Like Inversion, True/False elimination, DeMorgan, distributivity et cetera. I'ce come across a formula that's intuitively very clear, I just don't understand how to work it from A to B.

I'm asked to provide a calculation showing that: $P \wedge (P \lor Q) = P$. Intuitively I can see this to be true, drawing the truthtables confirms this, but how can I show this via a simple calculation?

I know I can rewrite this as follows: by distribution I get: $(P \wedge P) \lor (P \wedge Q) = P$, from which by idempotence follows that: $P \lor (P \wedge Q) = P$. Which actually leads me to exercise b . To show that is equivalent to P.

The big question is: how do I eliminate the Q? What rule can I apply here to show the equivalence to P?

2. ## Re: Showing equality via a simple calculation

$P\lor(P\land Q)=(P\land 1)\lor(P\land Q)=P\land(1\lor Q)=P\land1=P$.

3. ## Re: Showing equality via a simple calculation

Ah I see now, thanks! I haven't seen distributive and true/false elimination being used like that