From \(p \to (q \lor r)\): whenever p is true, q or r is true.

If q is true, then s is true (from \(q \to s\)). If r is true, then s is true (from \(r \to s\)).

So in either case s is true, which gives \(p \to s\). (This is the rule of constructive dilemma.)

Answer: (C) p → s