COT3100Logic02 - ¬ q ⇒ ¬ p 1 1 1 1 1 1 1 1 1 1 1 1 1 1...

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Rules of Inference 1) p This is the Rule of Detachment or Modus Ponens p q --------- q This can be proved by the truth table below. Very simply, given a premise and an implication using that premise, the conclusion of the implication must follow. p q p q [p (p q)] q 0 0 1 1 0 1 1 1 1 0 0 1 1 1 1 1
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2) p q This is the Law of Syllogism q r --------- p r Consider the logical form of this law: [(p q) (q r)] (p r) Essentially, this reduces to showing ¬ [(p q) (q r)] (p r) If p is false, we see the statement is true automatically. Thus we must only worry about the case where p is true. In this case, if r is true, we see we are fine as well. Thus, the final case to consider is if p is true and r is false. Regardless of q’s value at least one of (p q) and (q r) must be false in this situation, making the entire assumption true.
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3) p q This is Modus Tollens ¬ q --------- ¬ p In logical form we have [(p q) ¬ q] ¬ p Here is a truth table to verify this rule: p q ¬ p ¬ q p q [(p q)
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Unformatted text preview: ¬ q] ⇒ ¬ p 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 The rest of the rules of implication are listed on page 79 of the book. You will get more practice using these in recitation. Practice Problem : Using the rules of inference, and given the following premises: p ⇒ (q ⇒ r) p ∨ s t ⇒ q ¬ s Show that ¬ r ⇒ ¬ t must be true. Here is a list of the other rules stated in the text, without proof: p q-------p ∧ q (Rule of Conjuction) p ∨ q ¬ p-------q (Rule of Disjunctive Syllogism) ¬ p → F-----------p (Rule of Contradiction) p ∧ q-----------p (Rule of Conjunctive Simplification) p----------p ∨ q (Rule of Disjunctive Amplification) p ∧ q p → (q → r)----------------r (Rule of Conditional Proof) p → r q → r---------(p ∨ q) → r (Rule for Proof by Cases) p → q r → s p ∨ r---------q ∨ s (Rule of the Constructive Dilemma) p → q r → s ¬ q ∨ ¬ s-----------¬ p ∨ ¬ r...
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COT3100Logic02 - ¬ q ⇒ ¬ p 1 1 1 1 1 1 1 1 1 1 1 1 1 1...

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