COMP 232: Assignment 2
40165495 - Gabrielle Lemieux
Winter 2021 - Feb 26, 2021
Question 1 :
(a) Not equivalent
p = True; q = False; r = False
p⊕q⊕r
p ⊕ q ≡ True
(i) p ⊕ q ⊕ r ≡ True
p↔q↔r
p ↔ q ≡ False
(ii) p ↔ q ↔ r ≡ False
(i) and (ii) aren’t equivalent.
(b) Equivalent
∃(P (x) → Q(x))
≡ ∃x(¬P (x) ∨ Q(x))
≡ ∃x(¬P (x)) ∨ ∃xQ(x)
≡ ¬(∃xP (x)) ∨ ∃xQ(x)
∃xP (x) → ∃xQ(x)
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Question 2 :
(a) This statement would be true when there is an x for which (x = y + 3)
is true for every y. However, there is no x value for which (x = y + 3) is true
every y value within the domain. Therefore, this statement is false.
(b) This statement would be true when (x = y + 3) → (y > 0) is true for
every pair (x, y). In the case of the conditional, if the second premise is true,
the entire statement will be true every time. Since, the premise (y > 0) is
always true within the domain, this statement is true; for pair (x, y), the
statement P (x, y) is true.
(c) This statement would be true when there is an x for which (x = y + 3) →
(y < 0) is true for every y. When a conditional is present, if the second
premise is declared false, no matter the truth value of the first premise, the
whole statement is false. Furthermore, (y < 0), will always be false within
the domain. So, we can conclude that the statement is false.
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Question 3:
For (a) through (f):
- Capital(x, y): x is capital the of y
- Visit(x, y): x has visited y
- S(x): x is a student
(a) ∀x∃y(Capital(y, x) ∧ ∀z(Capital(z, x) → (y = z)))
(b) Capital(M ontreal, Canada)
(c) V isit(Jean, M ontreal) ∧ V isit(Jean, Ottawa)
(d) ∀(x)(S(x) → V isit(x, M ontreal))
(e) ∃(x)(V isit(x, y) ∧ V isit(x, z) ∧ y 6= M ontreal ∧ z 6= M ontreal ∧ y 6= z)
(f) ∀(x)¬∃(y)(S(x) → V (x, y))
For (g) through (h):
- F(x, y): x and y are students who are friends
- C(x, y): x is a student and they’re country of origin is y
(g) ∃(x)∃(y)¬∃(z)(F (x, y) ∧ F (x, z) ∧ F (y, z))
(h) ∃(x)∃(y)∃(z)∃(c)(C(z, c) ∧ (F (x, z) ∨ F (y, z)))
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Question 4:
(a) P (1, 3) ∨ P (2, 3) ∨ P (3, 3)
(b) ¬(P (2, 1) ∧ P (2, 2) ∧ P (2, 3))
(c) [P (1, 1)∨P (1, 2)∨P (1, 3)]∧[P (2, 1)∨P (2, 2)∨P (2, 3)]∧[P (3, 1)∨P (3, 2)∨
P (3, 3)]
(d) ¬[P (1, 1) ∨ P (1, 2) ∨ P (1, 3)] ∧ ¬[P (2, 1) ∨ P (2, 2) ∨ P (2, 3)] ∧ ¬[P (3, 1) ∨
P (3, 2) ∨ P (3, 3)]
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Question 5:
(a) Valid
1. p → r Premise 1
2. q → s Premise 2
3. ¬r ∧ q Premise 3
4. ¬r (3) Simplification
5. q (3) Simplification
6. s (2) and (5) Modus ponens
7. s ∨ r (6) Addition
(b) Invalid
1. p → (q → r) Premise 1
2. r → (p → q) Premise 2
3. q Premise 3
When all the premises are true, the outcome ¬ r, is false.
(c) Valid
1. (p → (q → (p ∧ q))) Premise 1
2. ¬p ∨ (q → (p ∧ q)) (1) Material Implication
3. ¬p ∨ ¬q ∨ (p) (2) Material Implication
4. ¬p(p) ∨ (p) (3) De Morgan Law
5. (p ∧ q) ∨ ¬(p ∧ q) (4) Commutativity
6. ¬(¬(p ∧ q)) ∨ (¬p ∨ ¬q) (5) Double Negation and Material Implication
7. ¬(¬(p ∧ q)) ∨ (p → ¬q) (6) Material Implication
8. ¬(p ∧ q) → (p → ¬q) (7) Material Implication
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(d) Valid
- P(x): x knows how to program
- C(x): x does well in Computer Science
- G(x): x has a job guaranteed
Domain: Everyone in this class
P (x) → C(x)
C(x) → G(x)
P (x) → G(x)
Follows Hypothetical Syllogism
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