2
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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.

  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.)

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

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2
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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
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  • $\begingroup$ You used conditional proof. $\endgroup$ – Doug Spoonwood Sep 11 '13 at 2:17
  • $\begingroup$ It is a Proof by Cases. $\endgroup$ – Graham Kemp Mar 25 at 22:36
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(1) PvP

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

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

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

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  • $\begingroup$ What's wrong with my proof? $\endgroup$ – Ray LittleRock Mar 25 at 19:38
  • 1
    $\begingroup$ It looks reasonable to me. $\endgroup$ – MJD Mar 25 at 20:21

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