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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