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Please use the discussion area to explain this fascinating question Constructive Dilemma Proof: 1. ( A -> B ) Premise 2. ( C -> D ) Premise 3. ( A or C ) Premise 4. ( - B -> - A ) 1 , Contraposition 5. ( - A -> C ) 3 , Implication 6. ( - B -> C ) 4 , 5 Chain Argument 7. ( - B -> D ) 6 , 2 Chain Argument 8. ( B or D ) 7 Implication , Q.E.D.

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Q: What is the proof of constructive dilemma in logic NOT with truth table if p is q and if r is s and either p or r is true therefore either q or s is true?
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