[(p or q) and (q implies r) and not r] implies (p and q) is false: truth values of p implies q and q implies p

The same question, asked 2 times: 11 Apr 2026 S1, 13 Apr 2026 S2. Every wording below, oldest first.

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Question 1
11 Apr 2026 S1
60 $[(p \vee q) \wedge (q \to r) \wedge (\sim r)] \to (p \wedge q)$
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Question 2
13 Apr 2026 S2
62 $[(p \vee q) \wedge (q \to r) \wedge (\sim r)] \to (p \wedge q)$
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