1. Discrete Structures
Lecture 11
Implication II 1

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 6

4. Modus Ponens
(3.77) Modus Ponens :
p  (p  q)  q 6

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

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) 6

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 6

8. Recall (1.5) Leibniz X = Y E[z:=X] = E[z:=Y]
Can be rewritten (notationally) as:
EzX = EzY
If X = Y is valid (true in all states), then so is E[z:=X] = E[z:=Y]. 6

9. Leibniz’s rule as an axiom
(3.83) Axiom, Leibniz :
(e = f)  Eze = Ezf
If e = f is true in a particular 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]. 6

10. Rules of Substitution that follow from Leibniz Axiom
(a) (e = f)  Eze  (e = f)  Ezf
(b) (e = f)  Eze  (e = f)  Ezf
(c) q  (e = f)  Eze
 q  (e = f)  Ezf 6

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. 6

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. 6

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 6

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
(E[z := p]  p) V q
= < (3.86a) E[z:=p]  p  E[z:=false]  p >
(E[z := false]  p) V q
¬E[z := false] V p V q
E[z := false]  p V q 6

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)
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 6