SOLUTION: Conditional Proof - can use all 18 rules (P -> Q) <-> (P -> (Q v ~P))

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Question 1009930: Conditional Proof - can use all 18 rules
(P -> Q) <-> (P -> (Q v ~P))
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Use a conditional proof twice

NumberStatementLines UsedReason
:.(P -> Q) <--> (P -> (Q v ~P))
1P -> QACP
2~P v Q1MI
3(~P v Q) v ~P2Add
4~P v (Q v ~P)3Assoc
5P -> (Q v ~P)4MI
6[P -> Q] -> [P -> (Q v ~P)]1-5CP
7P -> (Q v ~P)ACP
8~P v (Q v ~P)7MI
9~P v (~P v Q)8Comm
10(~P v ~P) v Q9Assoc
11~P v Q10Taut
12P -> Q11MI
13[P -> (Q v ~P)] -> [P -> Q]7-12CP
14{[P -> Q] -> [P -> (Q v ~P)]} & {[P -> (Q v ~P)] -> [P -> Q]}6,13Conj
15(P -> Q) <--> (P -> (Q v ~P))14ME



ACP = Assumption for Conditional Proof
Add = Addition
Assoc = Association
Comm = Commutation
Conj = Conjunction
CP = Conditional Proof
Dist = Distribution
DM = De Morgan's Law
HS = Hypothetical Syllogism
ME = Material Equivalence
MI = Material Implication
MP = Modus Ponens
Simp = Simplification
Taut = Tautology