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124MS
Logic and Sets
Logic and Sets: Coursework 1
Question One:
Find the truth value of X ∧ ((Y ⇒ W) ⇔ Z) if X is true, Y is true, and W is false and Z is false. What
is the truth value of this expression if the brackets are removed?
Answer:
X = T, Y = T, W = F, Z = F
T ∧ (( T ⇒ F ) ⇔ F )
T ∧ (( F ) ⇔ F )
T ∧ (( T ))
T∧T≡T
The truth value of the statement X ∧ ((Y ⇒ W) ⇔ Z) is true, as shown by my above workings.
Now I will attempt to work out the truth value of X ∧ Y ⇒ W ⇔ Z.
X ∧Y ⇒ W ⇔Z
T∧T⇒F⇔F
T⇒F⇔F
F⇔F≡T
The statement X ∧ Y ⇒ W ⇔ Z is also true.
Question Two:
Consider the following proposition:A ∧ (A ⇒ B) ∧ (B ⇒ ¬C) ⇒ B ∧ ¬C
(a) Interpreting these symbols as the atomic sentences
A : Logic is part of mathematics.
B : Mathematics is hard.
C : I enjoy logic.
Interpret the above proposition as an argument in natural English:
Answer:
Logic is part of mathematics and if logic is part of mathematics then mathematics is hard. Also, if
mathematics is hard then I do not enjoy logic, therefore mathematics is hard and I do not enjoy
logic.
124MS
Logic and Sets
(b) By means of a truth table, find out whether the argument is valid.
Answer:
A ∧ (A ⇒ B) ∧ (B ⇒ ¬C) ⇒ B ∧ ¬C
A
t
t
t
t
f
f
f
f
1
∧
t
t
f
f
f
f
f
f
10
(A
t
t
t
t
f
f
f
f
2
⇒
t
t
f
f
t
t
t
t
8
B)
t
t
f
f
t
t
f
f
3
∧
f
t
f
f
f
f
f
f
11
(B
t
t
f
f
t
t
f
f
4
⇒
f
t
t
t
f
t
t
t
9
¬C)
f
t
f
t
f
t
f
t
5
⇒
t
t
t
t
t
t
t
t
13
B
t
t
f
f
t
t
f
f
6
∧
f
t
f
f
f
t
f
f
12
¬C
f
t
f
t
f
t
f
t
7
The statement given, A ∧ (A ⇒ B) ∧ (B ⇒ ¬C) ⇒ B ∧ ¬C is a tautology because column 13 only
reveals true values, thus making the statement that is given valid.
Question Three:
Consider the following collection of statements:I slept in this morning. If I slept in this morning, and I walk to work, I will be late today. If the
buses are not running, I walk to work.
(a) Choose symbols to represent the atomic sentences in the above argument, and hence
express the statements in terms of propositional calculus.
Answer:
S : I slept in this morning
W : I walk to work
L : I will be late today
B : The buses are running
S ∧ ( S ∧ W ⇒ L ) ∧ ( ¬B ⇒ W )
124MS
Logic and Sets
(b) Construct a formal proof (table of assertions and justifications) showing that the statement
If I will not be late this morning, the buses are running follows from these hypotheses.
Answer:
Statement: S ∧ ( S ∧ W ⇒ L ) ∧ ( ¬B ⇒ W )
H1: S, H2: S ∧ W ⇒ L, H3: ¬B ⇒ W, H4: ¬L ⇒ B
Here I will construct a tableau of assertions and justifications.
Number
1
2
3
4
5
6
7
8
9
Assertion
¬L ⇒ B
¬L
S
S∧W⇒L
¬B ⇒ W
(S∧W)
¬S ∧ ¬W
¬W
B
Justification
H4
Deduction H4
H1
H2
H3
Modus Tolens, 2, 4
De Morgan’s, 4
Disjunctive Syllogism, 6, 7
Modus Tolens, 7, 8
So we see that B follows from H1, H2, H3, H4 and so from this, ¬L ⇒ B follows from H1, H2, H3, H4
by the deduction theorem.
(c) How many rows would be required by a truth table to show this?
Answer:
As there are 4 different symbols representing atomic sentences it would mean that 2 4, or 16,
rows will be required in the truth table to show the statements expressed in propositional
calculus.
124MS
Logic and Sets
Question Four:
I am cataloguing my collection of 25 toy cars. 12 of them are big, 9 are red, and 13 are scratched.
3 are big and red, 5 are big and scratched, and 3 are red and scratched. How many are red, but
not big or scratched?
Answer:
The universe is T; which represents Toy Cars. B represents Big toy cars. R represents Red toy
cars and S represents Scratched toy cars.
T
2
B
5
3
2
1
R
3
S
3
|R ∪ B ∪ S| = |R| + |B| + |S| - |R ∩ B| - |R ∩ S| - |B ∩ S| + |R ∩ B ∩ S|
25 = 9 + 12 + 13 – 3 – 3 – 5 + 2
R= 9–3 –3 –2
R= 1
Out of 25 toy cars, only 1 red car is not big or scratched.
124MS
Logic and Sets
Question Five:
(a) Use a hybrid truth/membership table to show that if A ⊆ B, then A ∪ C ⊆ B ∪ C.
Answer:
A
1
1
1
1
0
0
0
0
1
⊆
t
t
f
f
t
t
t
t
7
B
1
1
0
0
1
1
0
0
2
⇒
t
t
t
t
t
t
t
t
11
A
1
1
1
1
0
0
0
0
3
∪
t
t
t
t
t
f
t
f
8
C
1
0
1
0
1
0
1
0
4
⊆
t
t
t
f
t
t
t
t
10
B
1
1
0
0
1
1
0
0
5
∪
t
t
t
f
t
t
t
f
9
C
1
0
1
0
1
0
1
0
6
From column 11 being a tautology, we are shown that if A ⊆ B, then A ∪ C ⊆ B ∪ C.
(b) Give an example of a situation where the relationship A ∪ C ⊆ B ∪ C holds, even though A is
not a subset of B, where A, B and C are subsets of the universal set {1, 2, 3, 4}.
Answer:
A= {1 , 3}
B={2 ,3}
C= {1,2 }
Question Six:
a[1 . . . n] is an array of integers, and N is the set {1, 2, . . . , n − 1}.
(a) Express the statement ∀i ∈ N|a[i] < a[i + 1] in normal English.
Answer:
Everything in the array is increasing.
(b) Give the negation of this statement in formal language.
Answer:
¬∀i ∈ N|a[i] < a[i + 1]