Verified answer

★ “ if p then q” is equivalent to “(not p) or q”

I think the statement below is obviously true.

☆ “ if p then q” and “ if p´ then q´”　⇒　if（p or p´）then (q or q´)

Proof 1

(1) If P then [(Q and R) or S]

(2) If (Q and R) then Not P

(3) If T then Not S

...

Therefore, If P then Not T

【Solution】

From （3）, T→not S ⇔　not (not S) → not T ⇔ S → not T　….(4)

From (2) , (Q and R) → Not P ….(5)

From (4), (5) and ☆ , (Q and R) or S → (Not P) or (not T) ….(6)

From (1) ,(6) and ★, P → (Not P) or (not T) ⇔ (not P) or (not P) or (not T)　

⇔ (not P) or (not T) ⇔ P → (not T) ⇔　If P then Not T

Proof 2

(1) If A then B

(2) A or (B and C)

...

Therefore, B

【Solution】

From (2) and ★, (not A)→(B and C) ⇒ (not A) → B ….(3)

From (1) and (3) , { A or (not A) } → Ｂ

A or (not A) is always true.

Therefore B