Transcript Inference Rules: Tautologies •

Inference Rules: Tautologies
•(p  (p  q))  q.
•(q  (p  q))  p.
•p  (p  q).
•(p  q)  p.
•((p)  (q))  p  q.
•((p  q)  q)  p.
Modus Ponens
Modus Tollens
Addition
Simplification
Conjunction
Disjunctive Syllogism
•((p  q)  (q  r))  (p  r) Hypothetical Syllogism
• (aka, the transitive rule)
•((p  q )  ( p  r))  (q  r)
Resolution
•((p  q )  (p  r)  (q  r))  r Cases
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Fallacies
• (q  (p  q))  p.
•
Fallacy of Affirming the Conclusion
• (p  (p  q))  q.
•
Fallacy of Denying the Hypothesis
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Inference Rules: Quantified Statements
x P(x)
 P(c)
Universal Instantiation
P(c)___
 x P(x)
Universal Generalization
(for an individual object c from UoD)
(for any arbitrary element c from UoD)
x P(x)
 P(c)
Existential Instantiation
P(c)__
 x P(x)
Existential Generalization
(for some specific object c from UoD)
(for some object c from UoD)
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Proof Methods
• Trivial Proof of p  q given q (see truth table).
• Vacuous Proof of p  q given p (see truth table).
• Direct Proof of p  q given p, show q.
• Indirect Proof of p  q (proof by contraposition, so
• show q  p starting with q).
• Proof of p  q by Contradiction: recall that p  q Ξ
• p  q. Assume the negation, p  q, is true and
• show a contradiction.
• Proof of p1  p2  p3  … pn  q by Cases: show that
• pi  q is true for all i, where 1 ≤ i ≤ n.
• Proof of x P(x) by example: just find an instance.
• Proof that x P(x) is false by counterexample: just
• find an instance c where P(c) is false.
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