SOLUTION: 1. (A∨~B)→(F∨(R∙G)) 2. A 3. F→L 4. (R∙G)→T 5. (L∨T)→S ∴ S

Algebra ->  Proofs -> SOLUTION: 1. (A∨~B)→(F∨(R∙G)) 2. A 3. F→L 4. (R∙G)→T 5. (L∨T)→S ∴ S       Log On


   

Question 1116612: 1. (A∨~B)→(F∨(R∙G))
2. A
3. F→L
4. (R∙G)→T
5. (L∨T)→S ∴ S
Answer by math_helper(2461) About Me  (Show Source):
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1. (A∨~B)→(F∨(R∙G))   Premise

2. A                  Premise          

3. F→L                Premise

4. (R∙G)→T            Premise

5. (L∨T)→S            Premise

6. A v ~B             2  Addition (ADD)

7.  F v (R∙G)         6,1  Modus Ponens (MP)

8. L v T             7,3,4  Constructive Dilemma (CD)

9. S                  8,5  MP  (conclusion)



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After reviewing my first answer, I realized that the conditional proof was not only incorrect, it was unnecessary.