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Discrete Structures Lecture 11 Implication II

Discrete Structures Lecture 11 Implication II. Weakening and Strengthening. A formula is STRONGER if it restricts the state more. A formula is WEAKER when the fewest restrictions are in place. Example: p  q is true in only one state. p V q is true in three of four states.

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Discrete Structures Lecture 11 Implication II

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  1. Discrete Structures Lecture 11 Implication II

  2. Weakening and Strengthening A formula is STRONGER if it restricts the state more. A formula is WEAKER when the fewest restrictions are in place. Example: p  q is true in only one state. p V q is true in three of four states. Therefore, p  q is stronger. The formula true is the weakest (true in all states) The formula false is the strongest (true in no states).

  3. Weakening and Strengthening (3.76) Weakening/Strengthening : (a)p  p V q (b)p  q  q (c)p  q  p V q (d)p V (q  r)  p V q (e)p  q  p  (q V r) weakening: transform antecedent into the consequent strengthening: transform consequent into the antecedent

  4. Modus Ponens (3.77) Modus Ponens : p  (p  q)  q

  5. Case Analysis (3.78)(p  r)  (q  r)  (p V q  r) To prove (p V q  r) can prove(p  r) and (q  r)separately. (3.79)(p  r)  (¬p  r)  r Can prove r by breaking proof into two pieces.

  6. Mutual Implication (3.80) Mutual Implication : proof in text! (p  q)  (q p)  p  q (3.81) Antisymmetry : (p  q)  (q p)  (p  q)

  7. Transitivity (3.82) Transitivity: (a)(p  q)  (q r)  (p  r) Proof in text! (b)(p  q)  (q r)  (p  r) antecedent replacement (c)(p  q)  (q r)  (p  r) consequent replacement

  8. Recall (1.5) Leibniz X = Y E[z:=X] = E[z:=Y] Can be rewritten (notationally) as: X = Y EzX = EzY If X = Y is valid (true in all states), then so is E[z:=X] = E[z:=Y].

  9. Leibniz’s rule as an axiom (3.83) Axiom, Leibniz : (e = f)  Eze = Ezf If e = f is true in aparticular state, then so is E[z:= e] = E[z:=f] (in that state). which is different from saying if X = Y is valid (true in all states), then so is E[z:= X] = E[z:=Y].

  10. Rules of Substitution that follow from Leibniz Axiom (3.84) Substitution: (a) (e = f)  Eze (e = f)  Ezf (b) (e = f)  Eze (e = f)  Ezf (c) q  (e = f)  Eze  q  (e = f)  Ezf

  11. Replacing Variables by Boolean Constants (3.85) Replace by true : (a) p  Ezp p  Eztrue (b) q  p  Ezp q  p  Eztrue (a) any occurrence of the antecedent in the consequent can be replaced by true (b) extend to conjunction because both must be true.

  12. Replacing Variables by Boolean Constants (3.86) Replace by false : (a) Ezp p  Ezfalse p (b) Ezp q V p  Ezfalse q V p (a) replacing occurrences of the consequent in the antecedent. (b) extend to disjunction.

  13. Replacing Variables by Boolean Constants Continued (3.87) Replace by true : p  Ezp p  Eztrue (3.88) Replace by false : p V Ezp p V Ezfalse (3.89) Shannon : Ezp (p  Eztrue) V (¬p  Ezfalse) case analysis

  14. (3.86) Replace by false :(b) Ezp p V q  Ezfalse p V q Problem 3.79 says to prove (3.86b) E[z := p]  p V q = < (3.59) Implication > ¬E[z := p] V p V q = < (3.59) Implication > (E[z := p]  p) V q = < (3.86a) E[z:=p]  p  E[z:=false]  p > (E[z := false]  p) V q = < (3.59) Implication > ¬E[z := false] V p V q = < (3.59) Implication > E[z := false]  p V q

  15. (3.76) Weakening/Strengthening : (e)p  q  p  (q V r) Problem 3.84 says to prove (3.76e), using Replace by true (3.85b) p  q  p  (q V r) = < (3.85b) Replace by true > p  q  true  (q V r) = < (3.85b) Replace by true > p  q  true  (true V r) = < (3.29) Zero of V; (3.38) Idempotency of  (with p:=true)> p  q  true = < (3.72) Right Zero of  > true

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