# Showing equality via a simple calculation

• Oct 1st 2011, 01:22 AM
Lepzed
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: \$\displaystyle 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: \$\displaystyle (P \wedge P) \lor (P \wedge Q) = P\$, from which by idempotence follows that: \$\displaystyle 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?
• Oct 1st 2011, 02:11 PM
emakarov
Re: Showing equality via a simple calculation
\$\displaystyle P\lor(P\land Q)=(P\land 1)\lor(P\land Q)=P\land(1\lor Q)=P\land1=P\$.
• Oct 2nd 2011, 01:33 AM
Lepzed
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