SOLUTION: P v Q, P → (T → S), P → T, S ↔ Q ├ S
1. P v Q A
2. P → (T → S) A
3. P → T A
4. S ↔ Q A
5.
Algebra ->
Proofs
-> SOLUTION: P v Q, P → (T → S), P → T, S ↔ Q ├ S
1. P v Q A
2. P → (T → S) A
3. P → T A
4. S ↔ Q A
5.
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Question 1087187: P v Q, P → (T → S), P → T, S ↔ Q ├ S
1. P v Q A
2. P → (T → S) A
3. P → T A
4. S ↔ Q A
5. P 1 v O
6. Q 1 v O
7. S →Q 4 ↔ O
8. S 7, 6 → O
Do I have all of the steps completed?
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Using a proof by contradiction, we can say
The basic idea is to assume the opposite of the conclusion (~S) and prove that it leads to a contradiction, which it does on line 17. So the opposite of the assumption must be true, ie the conclusion (S) is true.
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