PHIL 110 Summer 2022 Assignment 3 - answer key Simon Fraser University PHIL 110

PHIL 110 – Introduction to logic and reasoning
Assignment 3

Distributed on Canvas: Wed. Jul. 27
Due at the beginning of class: Wed. Aug. 3
Value: 11.67%

Please read carefully. This assignment is out of 45 points (+2 bonus points) and is worth
11.67% of your final mark. On separate paper, answer each of the following questions. You
can type up or hand-write your work. If you write up your work by hand, please write
clearly. You must submit a hard copy of your work at the beginning of class. Electronic
submissions via email will not be accepted. Make sure to include your name, student ID,
and tutorial number on the front page of your assignment. Failure to do so will result in
a loss of marks.

Note: It should be remembered that the assignments are not just assessment tests. They also
have a pedagogical function, in that they’re designed to push you to look into things which
you might not already have looked at sufficiently closely.

Question about Language (10 pts.)
Suppose we have an FOL with constants a, b, c, d, e, variables x, y, z, predicate symbols Px,
Qx, Rxy, x = y, the regular connectives ¬, ∧,∨, →,↔, and the quantifiers ∀ and ∃. For each
of the following, indicate whether they are simple terms, mere well-formed formulas
(wffs), atomic sentences, complex (non-atomic) sentences, or none of these.

a. ∀x((Rba ∧ Qec) → ∃y(y = x ↔ Pd)) f. ∀x(Pa → Qa)
None Complex sentence
b. [∃x∃y(Px ∧ Rxa) ∧ Qx] g. e
Mere wff Simple term
c. ∀a (Pa → Qa) h. a = c
None atomic sentence
d. Rax i. (a = c ∧ b = c)
Mere wff Complex sentence
e. ∃z (Qz ∧ Rad) j. ∀x∃x(Lx → Txx)
Complex sentence None


Translations (10 pts.)
Consider the following symbolization key:

UD: Everything
Cxy: x cares for y
Oxy: x owns y
Px: x is a pet
Sx: x is a student


1

, 1. (5 pts.) Translate the following English sentences in FOL:
i. Some students have no pet.
∃x(Sx ∧ ∀y(Py → ¬Oxy))

ii. No student owns every pet.
∀x(Sx → ¬∀y(Py → Oxy))

iii. Some pets aren’t owned by any student.
∃x(Px ∧ ∀y(Sy → ¬Oyx)

iv. Every student who owns a pet cares for it.
∀x(Px → ∀y((Sy ∧ Oyx) → Cyx))

v. Every pet is owned and cared for by some student.
∀x(Px → ∃y((Sy ∧ Oyx) ∧ Cyx))

2. (5 pts.) Given the symbolization key above, translate the following FOL sentences in
English:
i. ∀x(Px → ∀y(Oyx → Cxy))
Every pet cares for each of their owners.

ii. ∃x(Px ∧ ∃y(Oyx ∧ ¬Cyx))
Some pet isn’t cared for by their owner.

iii. ∀y(Sy → ∀z(Pz → ¬Oyz))
No student owns any pet.

iv. ∀x(Sx → ¬∀y(Py → Oxy))
No student owns every pet.

v. ∃x(Px ∧ ∃y((Sy ∧ Oyx) ∧ ¬Cxy))
Some pet doesn’t care for some of their student owner.


Explanation (5 pts.)

Show that ∨E is sound in t.

For the sake of contradiction, suppose that there is a proof p in t whose first invalid step
derives the sentence P from an application of ∨E to sentences (P ∨ Q) and ¬Q. Let A1…Ak,
(P ∨ Q), and ¬Q be the assumptions in force at step P. Since we’re assuming that this is the
first invalid step, we’re assuming that P is not a tautological consequence of A1…Ak, (P ∨ Q),
and ¬Q. Now consider the joint truth table for A1…Ak, (P ∨ Q), ¬Q, and P. Since we’re
assuming that P is not a tautological consequence of A1…Ak, (P ∨ Q), and ¬Q, there must be

2