# (Symbolic Logic) Proving P v P = P (Idempotency) using a direct proof

1. Sep 10, 2013

### jdinatale

Ok, so it's very easy to show P v P = P (where = is logically equivalent) using a truth table as well as using a conditional proof.

P v P Premise
~p Assumption
p Disjunctive Syllogism (1, 2)
p & ~p Conjunction (3, 4)
~p --> (p & ~p) Conditional Proof (2--4)
p v ~p EMI
~p v p Commutation (6)
~p v ~~p Double Negation (7)
~(p & ~p) De Morgan's (8)
~~p Modus Tollens (5, 9)
p Double Negation

My question is, how do I show p v p = p WITHOUT using a truth table OR a conditional prove? I can only use the basic rules of inference (Excluded Middle Introduction, Disjunctive Syllogism, Addition, Conjunction, Simplification) as well as the rules of replacement (De Morgan's, Distribution, etc.)

2. Sep 11, 2013

### CompuChip

The <= direction, P |- P v P, is trivial by addition. The tricky direction is =>: P v P |- P. I think you would need at least disjunctive elimination ({P |- R, Q |- R, P v Q} |- R) for that so if that's not in your basic set you should probably try and derive it.

3. Sep 11, 2013

### jdinatale

I'm not allowed to do that in this class. I have to do a line by line prove and I can't use an implication. I see what you're doing.

I have to do something like this:

1. P v P........Premise
2. P v P v ~P......Addition
3. P v ~(~P v P) De Morgans

etc.