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

  • Thread starter jdinatale
  • Start date
  • #1
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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.)
 

Answers and Replies

  • #2
CompuChip
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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
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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.
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.
 

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