Question

construct the truth table (p ∧ q) =⇒ [(q ∧ ¬p) =⇒ (r ∧ q)]​

Answer 1

$\begin{array}{c|c|c|c|c|c} p & q & r & p\land q & q\land \neg p & r \land q \\&&&&\\ T & T & T & T & F & T \\ T & T & F & T & F & F \\ T & F & T & F & F & F \\ T & F & F & F & F & F \\ F & T & T & F & T & T \\ F & T & F & F & T & F \\ F & F & T & F & F & F \\ F & F & F & F & F & F\end{array}$

An implication A => B is true if either A is false, or both A and B are true. So

$\begin{array}{c|c|c}p\land q & (q\land\neg p) \implies (r\land q) & (p\land q) \implies \big[(q\land\neg p) \implies (r\land q)\big] \\&&\\T & T & \mathbf T\\T & T & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\\F & F & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\end{array}$

and the given statement is a tautology.