Download Logic and Sets: Coursework 1

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
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]