Question 1009333: Derive:(~P) v Q
1. P>Q
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! I'm going to do a proof by contradiction. It is also known as an indirect proof. I'm going to assume the complete opposite of what I want to derive. Then I'm going to show there's a contradiction that arises because of this assumption. Because of this contradiction, the complete opposite of the assumption must be true.
The assumption is made on line 2. It is the complete opposite of ~P v Q
The contradiction happens on line 9.
| Number | Statement | Lines Used | Reason |
|---|
| 1 | | P -> Q | | | | :. | | ~P v Q | | | | 2 | ~(~P v Q) | | AIP | | 3 | ~~P & ~Q | 2 | DM | | 4 | P & ~Q | 3 | DN | | 5 | ~Q & P | 4 | Comm | | 6 | P | 4 | Simp | | 7 | ~Q | 5 | Simp | | 8 | Q | 1,6 | MP | | 9 | Q & ~Q | 8,7 | Conj | | 10 | | ~P v Q | 2-9 | IP |
Acroynyms/Abbreviations used. I'm using the rules of inference/replacement
AIP = assumption for indirect proof (aka proof by contradiction)
Comm = commutation
Conj = conjunction
IP = indirect proof (aka proof by contradiction)
DM = de morgan's law
DN = double negation
MP = modus ponens
Simp = simplification
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