Transcript p  q

Discrete Structures
Lecture 11
Implication II
1
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).
2
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
3
Modus Ponens
(3.77) Modus Ponens :
p  (p  q)  q
4
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.
5
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
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
7
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].
8
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].
9
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
10
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.
11
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.
12
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 :
case analysis
Ezp  (p  Eztrue) V (¬p  Ezfalse)
13
(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
14
(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
15