If the statement q ∧ r is true, determine all combinations of truth values for p and s such that the statement (q → [¬p ∨ s]) ∧ [¬s → r] is true.

If the statement $q ∧ r$ is true, then $q$ is true and $r$ is true. Since $(q → [¬p ∨ s]) ∧ [¬s → r]$ is true, we conclude that $q → [¬p ∨ s]$ is true and $¬s → r$ is also must be true. Taking into account that $q → [¬p ∨ s]$ is true and $q$ is true, we conclude that $¬p ∨ s$ is also must be true. Since $r$ is true, the implication $¬s → r$ is true for any truth value of $s$. Taking into account that $¬p ∨ s$ is false if and only if $p$ is true and $s$ is false, we conclude that $(q → [¬p ∨ s]) ∧ [¬s → r]$ is true if and only if $p$ is true and $s$ is true or $p$ is false and $s$ is false or $p$ is false and $s$ is true.