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A Note of variable substitution
In each of the formulas I have gone over in chapters 2 and 3,
each variable is exactly that, a variable. A variable is
something that you can substitute in any value of the similar
type for.
..that is what makes formulas with variables so
powerful – they are applicable to a large number of cases. Here
is an example of what I mean:
The Distributive Law is:
p
∨
(q
∧
r)
⇔
(p
∨
q)
∧
(p
∨
r)
Imagine simplifying the following expression:
((s
∧
t)
∨
v)
∧
(
¬
v
∨
(s
∧
t))
((s
∧
t)
∨
v)
∧
((s
∧
t)
∨
¬
v), Commutative
(s
∧
t)
∨
(v
∧
¬
v),
Distributive, with p= s
∧
t, q= v, r=
¬
v
(s
∧
t)
∨
F,
Inverse Law
(s
∧
t) ,
Identity Law
One other thing to mention: notice how I used different letters
in the problem than are listed in the rules.
..The reason I did
that was for clarity’s sake; to show you all what was
substituted for each variable. BUT, in the problems that YOU
do, you’ll find that ps, qs and rs are used, just like in the
formula. When you make substitutions, you’ll have to
differentiate between the p&qs from the formula, and the
p&qs in your specific problem!!! Be careful when doing this.
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following:
¬
(
¬
((A
∪
B)
∩
C)
∪
¬
B) = (B
∩
C)
One way is to use a membership table:
A
B
C
B
∩
C
¬
((A
∪
B)
∩
C)
¬
B
¬
(
¬
((A
∪
B)
∩
C)
∪
¬
B)
0
0
0
0
1
1
0
0
0
1
0
1
1
0
0
1
0
0
1
0
0
0
1
1
1
0
0
1
1
0
0
0
1
1
0
1
0
1
0
0
1
0
1
1
0
0
1
0
0
1
1
1
1
0
0
1
Next, we could use Set Laws to show the two expressions to be
equivalent:
¬
(
¬
((A
∪
B)
∩
C)
∪
¬
B)
=
¬¬
(( A
∪
B)
∩
C)
∩
¬¬
B
(De Morgan’s)
= (( A
∪
B)
∩
C)
∩
B
(Double Negation)
= (( A
∪
B)
∩
B)
∩
C
(Associate &
Commutative)
= B
∩
C
(Absorption)
For this particular problem, we could show that each set is a
subset of the other. However, for this particular problem, that
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 Spring '09

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