Consider the following two propositions:
\[P1: \sim (p \to \sim q)\]
\[P2:(p \wedge \sim q) \wedge (( \sim p) \vee q)\]
If the proposition \[p \to (( \sim p) \vee q)\] is evaluated as FALSE, then :
Solutions · 1
Official solution
Given p→(∼p∨q) is false
⇒∼p∨q is false and p is true
Now p= True.
∼T∨q=FF∨q=F⇒q is false
P1:∼(T→∼F)≡∼(T→T)≡ False.
\[
P2:(T∧∼F)∧(∼T∨F)≡(T∧T)∧(F∨F)≡T∧F≡ False
\]