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If p and q are statements, then "$$\sim p \Rightarrow p \wedge q$$" is true whenever
If p and q are statements, then "$$\sim p \Rightarrow p \wedge q$$"
let solve that through the navigation of "$$\sim p \Rightarrow p \wedge q$$"
p→(p∨∼q)
−(p→(p∨∼q))
∵ − (p → q) = p∧ ∼ q
= p∧ ∼ ( p∨ ∼ q)= p∧(∼ q∧ ∼ q) = (p∧∼p)∧(p∧q)=t∧ ( p ∧ q) =t Answer
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