Among the statements
(S1) : \[(p \Rightarrow q) \vee((\sim p) \wedge q)\] is a tautology
(S2) : \[(q \Rightarrow p) \Rightarrow((\sim p) \wedge q)\] is a contradiction
Solutions · 1
Official solution
(S1) : \[(p \Rightarrow q) \vee((\sim p) \wedge q)\]
\[
\[\begin{array}{|cccccc|}\\ \mathrm{P} & \mathrm{Q} & \sim p & \sim p \wedge q & p \Rightarrow q & \begin{array}{c}
(p \Rightarrow q) \vee \\ (\sim p \wedge q)\end{array} \\ \\ \mathrm{T} \& \mathrm{T} \& \mathrm{F} \& \mathrm{F} \& \mathrm{T} \& \mathrm{T} \\ \\ \mathrm{T} \& \mathrm{F} \& \mathrm{F} \& \mathrm{F} \& \mathrm{F} \& \mathrm{F} \\ \\ \mathrm{F} \& \mathrm{T} \& \mathrm{T} \& \mathrm{T} \& \mathrm{T} \& \mathrm{T} \\ \\ \mathrm{F} \& \mathrm{F} \& \mathrm{T} \& \mathrm{F} \& \mathrm{T} \& \mathrm{T} \\ \\ \end{array}\]
\]
Here, for every value of components compound statement (last column) are not true. Hence, S1 is not a tautology.
S2:(q⇒p)⇒((∼p)∧q)