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

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.

1. P v P Premise
2. ~p Assumption
3. p Disjunctive Syllogism (1, 2)
4. p & ~p Conjunction (3, 4)
5. ~p --> (p & ~p) CP (2--4)
6. p v ~p EMI
7. ~p v p Commutation (6)
8. ~p v ~~p Double Negation (7)
9. ~(p & ~p) De Morgan's (8)
10. ~~p Modus Tollens (5, 9)
11. 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 (EMI, Disjunctive Syllogism, Addition, Conjunction, Simplification) as well as the rules of replacement (De Morgan's, Distribution, etc.)

• which rules are allowed? – Willemien Sep 10 '13 at 9:21
• The basic rules of inference (EMI, Disjunctive Syllogism, Addition, Conjunction, Simplification) as well as all the rules of replacement (De Morgan's, Distribution, etc.) – Joseph DiNatale Sep 10 '13 at 12:40

## 3 Answers

This looks like a Copi exercise, so I'll use the rules in the 1998 edition of "Introduction to Logic".

1 [(p $\lor$ p)=(p $\lor$ p)] $\lor$ commutation

2 [(p $\lor$ p)=$\lnot$$\lnot(p \lor p)] 1 Double Negation 3 [(p \lor p)=\lnot(\lnotp \land \lnot p)] 2 De Morgan's ahem... Petrus Hispanus's Theorems 4 [(p \lor p)=\lnot$$\lnot$p] 3 $\land$ tautology

5 [(p $\lor$ p)=p] 4 Double Negation

A normal proof would be:

1. P v P     Premise
2. |_ p      Assumption
3. |  P      2 reiteration
4  p         1,2,3,2,3 v Elimination

• You used conditional proof. – Doug Spoonwood Sep 11 '13 at 2:17
• It is a Proof by Cases. – Graham Kemp Mar 25 at 22:36

(1) PvP

(2) <=> ~ (~P & ~P) ______________ By definition of OR : (XvY) <=> ~ ( ~X & ~Y)

(3) <=> ~ ~P _____________________ By : idempotency of &

(4) <=> P _______________________ By : double negation

• What's wrong with my proof? – Ray LittleRock Mar 25 at 19:38
• It looks reasonable to me. – MJD Mar 25 at 20:21