SOLUTION: Use natural deduction to derive the conclusion in each problem. Use natural deduction to prove the following logical truth: (P ⊃ Q) ≡ [P ⊃ (Q ∨ ∼P)]

Question 1205226: Use natural deduction to derive the conclusion in each problem.

Use natural deduction to prove the following logical truth:
(P ⊃ Q) ≡ [P ⊃ (Q ∨ ∼P)]
Found 2 solutions by math_helper, Bogz:
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!

1. (P --> Q) == P --> (Q v ~P)
// Note to prove A == B, need to show A --> B and B --> A
// Also note I'm using simplified notation as it is easier to enter
2. :: P Conditional Proof (CP) assumption #1
3. :: Q 2,1 Modus Ponens (MP)
4. :: Q v ~P 3 Addition (ADD)
5. :: P --> (Q v ~P) 2-4 CP
6. :: (P --> Q) --> (P --> (Q v ~P)) 1-5 CP
// Now go the other way
7. :: P --> (Q v ~P) CP assumption #2
8. :: P CP assumption #3
9. :: (Q v ~P) 8,7 MP
10.:: P --> Q 9 Material Implication (MI)
11.:: (P --> (Q v ~P)) --> (P --> Q) 7-10 CP
12.:: (P --> Q) == (P --> (Q v ~P)) 6,11 Material Equivalence (ME)
13.(P --> Q) == (P --> (Q v ~P)) 2-12 CP
There may be shorter ways to go, but this is what came to mind.
Answer by Bogz(13) (Show Source): You can put this solution on YOUR website!
P ⊃ (Q ∨ ∼P)
= ~P V (Q ∨ ∼P) (equivalence to implication)
= ~P V (~P ∨ Q) (commutativity)
= (~P V ~P) ∨ Q (associativity)
= ~P V Q (simplification)
= P ⊃ Q (equivalence to implication)
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