SOLUTION: 1. Q > ~Q 2. ~(G & Q) > U / ~(G & ~U) Can somebody solve this proof please? Thank you :) "You can do a proof by contradiction. Step 1) assume the complete opposite of th

Question 981975: 1. Q > ~Q
2. ~(G & Q) > U
/ ~(G & ~U)
Can somebody solve this proof please? Thank you :)
"You can do a proof by contradiction.
Step 1) assume the complete opposite of the conclusion. Assume (G & ~U)
Step 2) Use simplification to get ~U
Step 3) Use modus tollens with line 2 and ~U to get ~~(G & Q) which turns into (G & Q)
Step 4) Simplification frees up Q
Step 5) Modus ponens on line 1, and using the freed up Q from step 4, consequently frees up ~Q
Step 6) We have Q and ~Q they conjunct to (Q & ~Q) which is always false. This is a contradiction.
Since we have a contradiction, the initial assumption (G & ~U) is false which makes the opposite true. That proves ~(G & ~U) is a proper conclusion."
Jim, is there anyway to do this proof without using a contradiction and not assuming? Thank you :)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Alternative method

Number	Statement	Lines Used	Reason
1	Q > ~Q
2	~(G & Q) > U
:.	~(G & ~U)
3	~Q v ~Q	1	Material Implication
4	~Q	3	Tautology
5	~~(G & Q) v U	2	Material Implication
6	(G & Q) v U	5	Double Negation
7	U v (G & Q)	6	Commutation
8	(U v G) & (U v Q)	7	Distribution
9	(U v Q) & (U v G)	8	Commutation
10	U v Q	9	Simplification
11	U	10,4	Disjunctive Syllogism
12	U v ~G	11	Addition
13	~G v U	12	Commutation
14	~(~~G & ~U)	13	De Morgan's Law
15	~(G & ~U)	14	Double Negation

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