SOLUTION: Very confused with this proof, any help would be appreciated!! Conditional Proof - can use all 18 rules P -> (~P -> (Q <-> (R v S)))

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Question 1009931: Very confused with this proof, any help would be appreciated!!
Conditional Proof - can use all 18 rules
P -> (~P -> (Q <-> (R v S)))

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
Conditional Proof

Number Statement Lines Used Reason
:. P -> (~P -> (Q <-> (R v S)))
1 P ACP
2 ~~P 1 DN
3 ~~P v (P -> (Q <-> (R v S))) 2 Add
4 ~P -> (P -> (Q <-> (R v S))) 3 MI
5 (~P & P) -> (Q <-> (R v S)) 4 Exp
6 (P & ~P) -> (Q <-> (R v S)) 5 Comm
7 P -> (~P -> (Q <-> (R v S))) 6 Exp
8 P -> {P -> (~P -> (Q <-> (R v S)))} 1-7 CP
9 (P & P) -> (~P -> (Q <-> (R v S))) 8 Exp
10 P -> (~P -> (Q <-> (R v S))) 9 Taut

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Proof by contradiction (alternate method)

Number Statement Lines Used Reason
:. P -> (~P -> (Q <-> (R v S)))
1 ~[P -> (~P -> (Q <-> (R v S)))] AIP
2 ~[~P v (~P -> (Q <-> (R v S)))] 1 MI
3 ~[~P v (~~P v (Q <-> (R v S)))] 2 MI
4 ~[~P v (P v (Q <-> (R v S)))] 3 DN
5 ~[(~P v P) v (Q <-> (R v S))] 4 Assoc
6 ~(~P v P) & ~(Q <-> (R v S)) 5 DM
7 ~(~P v P) 6 Simp
8 ~~P & ~P 7 DM
9 P & ~P 8 DN
10 P -> (~P -> (Q <-> (R v S))) 1-9 IP

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Abbreviations/Acronyms Used

ACP = Assumption for Conditional Proof
AIP = Assumption for Indirect Proof (aka proof by contradiction)
Add = Addition
Assoc = Associative Rule
Comm = Commutation
CP = Conditional Proof
DM = De Morgan's Law
DN = Double Negation
Exp = Exportation
IP = Indirect Proof (aka proof by contradiction)
MI = Material Implication
Simp = Simplification
Taut = Tautology