Question 1041461: Hello, can you please help me with this problem?
Use one of the indirect proof techniques (reductio ad absurdum or conditional proof) to demonstrate the validity of the argument.
~S → (F → L), F → (L → P), therefore, ~S → (F → P)
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
I'm going to use a conditional proof
| Number | Statement | Lines Used | Reason |
|---|
| 1 | | ~S -> (F -> L) | | | | 2 | | F -> (L -> P) | | | | :. | | ~S -> (F -> P) | | | | | | 3 | ~S | | ACP | | | | 4 | F -> L | 1,3 | MP | | | | 5 | (F & L) -> P | 2 | Exp | | | | 6 | (L & F) -> P | 5 | Comm | | | | 7 | L -> (F -> P) | 6 | Exp | | | | 8 | F -> (F -> P) | 4,7 | HS | | | | 9 | (F & F) -> P | 8 | Exp | | | | 10 | F -> P | 9 | Taut | | 11 | | ~S -> (F -> P) | 3-10 | CP |
Abbreviations/Acronyms Used:
ACP = Assumption for Conditional Proof
CP = Conditional Proof
Comm = Commutation
Exp = Exportation
HS = Hypothetical Syllogism
MP = Modus Ponens
Taut = Tautology
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