SOLUTION: Indirect proof
9.
1) R
2) (~ C v ~ D) v S
3) ~ (C ⋅ D) ⊃ ~R / ∴ S
10.
1) (A ⋅ B)⋅ ~ (S v T)
2) ~E
3) (S v T) v ~ (~E ⋅ ~F)
4) (~E
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Proofs
-> SOLUTION: Indirect proof
9.
1) R
2) (~ C v ~ D) v S
3) ~ (C ⋅ D) ⊃ ~R / ∴ S
10.
1) (A ⋅ B)⋅ ~ (S v T)
2) ~E
3) (S v T) v ~ (~E ⋅ ~F)
4) (~E
Log On
Question 1112710: Indirect proof
9.
1) R
2) (~ C v ~ D) v S
3) ~ (C ⋅ D) ⊃ ~R / ∴ S
10.
1) (A ⋅ B)⋅ ~ (S v T)
2) ~E
3) (S v T) v ~ (~E ⋅ ~F)
4) (~E v F)⊃(A ⋅ B) / ∴ E v F
12.
1) A v B
2) B ⊃ (A v D)
3) ~ D / ∴ A
Conditional proof
11.
1) A ⊃ B / ∴ A ⊃ [ C ⊃ ~ (B ⊃ ~A) ]
Help me please! Answer by solver91311(24713) (Show Source):
You can format this however you want but the basic idea is to assume the negation of the desired conclusion and have that lead you to a contradiction, that is a statement that is identically false.
Assume ~S. Then from (2), you get ~C v ~D by Disjunctive Syllogism. Then you can write ~(C & D) from the previous statement using De Morgan. From this and (3) you get ~R by Modus Ponens. But that leads to R & ~R which is identically false. Hence the assumption, ~S, is false, therefore S.
One question per post in accordance with the posting instructions, please.
John
My calculator said it, I believe it, that settles it
