SOLUTION: 1. Q > ~Q 2. ~(G & Q) > U / ~(G & ~U) Can somebody solve this proof please? Thank you :)

Question 981965: 1. Q > ~Q
2. ~(G & Q) > U
/ ~(G & ~U)
Can somebody solve this proof please? Thank you :)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
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.
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