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 |
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| 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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