Handout #51
CS103
May 18, 2010
Robert Plummer
Midterm Solutions
1.
Logic Proof (20 points)
(a)
In this part, you are proving one of the De Morgan theorems for quantifiers.
You are
not allowed to use (nor will you need) the other De Morgan theorems for quantifiers.
1.
¬
x P(x)
2.
Let c be an arbitrary member of the domain
3.
P(c)
4.
x P(x)
Existential generalization, 3
5.
Contradiction, 1, 4
6.
P(c)
Negation elimination, 35
7.
x ¬P(x)
Universal generalization, 26
That is all there is to it!
To prove that something is universally true [in this case ¬P(x)],
we consider an arbitrary element of the domain and show that the thing we want to prove
is true for that element.
So the form is the subproof from line 2 to line 6.
That is the only
way we can do a universal generalization, and the universally quantified result is outside
thesubproof.
Here is a common erroneous proof:
1.
¬
x P(x)
2.
¬
x ¬P(x)
3.
Let c be an arbitrary member of the domain
4.
¬P(c)
5.
x ¬P(x)
Universal generalization, 4
ERROR!
6.
Contradiction, 2, 5
7.
P(c)
Negation introduction, 46
8.
x P(x)
Existential introduction, 7
ERROR!
9.
Contradiction, 1, 8
10.
x ¬P(x)
Negation introduction, 29
You cannot do the universal generalization at step 5.
The pattern is not
the one shown in
the first proof.
Step 8 is also incorrect.
The options here would have been the existential
introduction inside the subproof, or universal introduction outside the subproof
(because
at this point you would
have the right pattern).
Neither helps.
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2
Here is another incorrect proof:
1.
¬
x P(x)
2.
¬
P(c)
Existential instantiation, 1
ERROR!
3.
x ¬P(x)
Universal generalization, 2
It is not possible to instantiate the existential quantifier, since it has a negation in front.
To do the instantiation, the first symbol has to be
.
You could get line 2 by first
applying De Morgan's Law, but that is what you are tying to prove!
(b) In this part, you
are allowed to use De Morgan's laws freely, both the quantified
and unquantified versions
, but cite them as justification if you do.
Note that 'b' is a
constant
in this proof.
1.
x ( (A(x)
¬B(x))
C(b) )
2.
¬
x C(x)
3.
y B(y)
4.
B(d)
For some d, existential instantiation, 3
5.
(A(d)
¬B(d))
C(b)
Universal instantiation, 1
6.
¬A(d)
¬B(d)
C(b)
Table 7, alternate form of
7.
¬A(d)
C(b)
Disjunctive syllogism, 4, 6
8.
x ¬C(x)
De Morgan's Law, 2
9.
¬C(b)
Universal instantiation, 8
10. ¬A(d)
Disjunctive syllogism,
7, 9
11.
x ¬A(x)
Existential instantiation, 10
12.
x A(x)
De Morgan's Law, 11
Most of you did well on this part.
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 Fall '09
 Logic, Mathematical Induction, Inductive Reasoning, Predicate logic, Quantification, Structural induction

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