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

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

-------------------------------------------------------------------------------

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

-------------------------------------------------------------------------------

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
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