SOLUTION: Help me solve the following using an indirect proof. I am including what I have so far. 1.(AvB)>(C&D) 2.(CvE)>~B 3.(DvF)>~A / ~A&~B 4.~(~A&~B) AIP 5.~~Av~~B 4,DM 6

SOLUTION: Help me solve the following using an indirect proof. I am including what I have so far. 1.(AvB)>(C&D) 2.(CvE)>~B 3.(DvF)>~A / ~A&~B 4.~(~A&~B) AIP 5.AvB 4,DM 6

Question 1030232: Help me solve the following using an indirect proof. I am including what I have so far.
1.(AvB)>(C&D)
2.(CvE)>~B
3.(DvF)>~A / ~A&~B
4.~(~A&~B) AIP
5.~~Av~~B 4,DM
6.C&D 1,5 MP
7.C 6, Simp

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

Number		Statement	Lines Used	Reason	Notes
1		(A v B) > (C & D)
2		(C v E) > ~B
3		(D v F) > ~A
:.		~A & ~B
	4	~(~A & ~B)		AIP
	5	~~A v ~~B	4	DM
	6	A v B	5	DN
	7	C & D	1,6	MP
	8	C	7	Simp
	9	D	7	Simp
	10	C v E	8	Add
	11	~B	2,10	MP
	12	D v F	9	Add
	13	~A	3,12	MP
	14	~A & ~B	13,11	Conj
	15	[~(~A & ~B)] & [~A & ~B]	4,14	Conj
16		~A & ~B	4-15	IP	See note below

Note: Line 4 and line 14 contradict one another. So IF you make the assumption ~(~A & ~B) (as done on line 4) then it leads to a contradiction (on line 14). That invalidates the iniatial assumption making the opposite of the assumption true. The opposite of ~(~A & ~B) is ~~(~A & ~B) which turns into ~A & ~B

Abbreivations Used:

Add: Addition
AIP: Assumption for Indirect Proof
Conj: Conjunction
DM: De Morgan's Law
DN: Double Negation
IP: Indirect Proof (aka proof by contradiction)
MP: Modus Ponens
Simp: Simplification
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