Fall 2015, Philosophy 209 Sample Test III with
Solutions
Dr. Holmes
December 3, 2015
The test will begin at 10:30 and end officially at 11:20: I will actually give
a five minute warning at that point. A summary of the rules is attached,
including the quantifier rules.
You are allowed use of your test paper and your writing instrument. Cell
phones should be turned off and out of sight.
General remarks about proofs: all propositional logic rules (including CP
and RAA) may be used in any proof. You should not be certain that every
premise is actually needed in every argument.
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1. Proofs of theorems (arguments without premises)
Write a proof of the theorem
(P → Q) → ¬(P • ¬Q)
Hint: since you have no premises, use CP and/or RAA to introduce
some!
Assume (1) P → Q for CP
Assume (2) P • ¬Q for RAA
(3) P simp (2)
(4) ¬Q simp (3)
(5) Q mp (1)(3)
(6) Q • ¬Q conj (4)(5)
(7) ¬(P • ¬Q) RAA 2-6
(8) (P → Q) → ¬(P • ¬Q) CP 1-8
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2. Supply line justifications in the proofs which follow.
The first example
1 (∃x)(P x) premise
2 (∀y.P y → Qy) premise
Goal: (∃z)(Qz))
3 P a EI 1
4 P a → Qa UI 2
5 Qa MP 3,4
6 (∃z)(Qz) EG 5
The second example
1 (x)(P x ∨ Qx) premise
2 (x)(P x → Rx) premise
Goal: (x)(Qx ∨ Rx)
3 P a ∨ Qa UI 1
4 Assume Qa for CP
5 Qa → Qa CP 4-4
6 P a → Ra UI 2
7 Qa ∨ Ra constructive dilemma 3,6,5
8 (x)(Qx ∨ Rx) UG 7 a introduced by UI, appeared in CP assumption
on line 4 but this was closed in line 5
3
3. Identify invalid steps in proofs (give reasons when you say a line is
invalid; you may just check a line which is valid without comment)
First example
1 (∃x)(P x)
2 (x)(P x → Qx)
Goal: (x)(Qx)
3 P a EI 1 valid
4 P a → Qa UI 2 valid
5 Qa MP 3,4 valid
6 (x)(Qx) UG 5 invalid, a is introduced by EI
Second example
1 (x)(P x) → (x)(Qx)
2 Pa
3 (x)(Rx ∨ ¬Qx)
Goal: Ra
4 P a → Qa UI 1 invalid, 1 is not a universally quantified statement
5 Qa MP 2,4
6 Ra ∨ ¬Qa UI 3 valid
7 Ra DS 5,6 not quite right, a dn step skipped (needed ¬¬Qa)
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4. Proof with just UI and EG
Prove that the following argument is valid.
1 (x)(P x → Qx)
2 (x)(Rx ∨ ¬Qx)
3 Pc
Goal: (∃w)(Rw)
(4) P c → Qc UI 1
(5) Rc ∨ ¬Qc UI 2
(6) Qc MP 3,4
(7) ¬¬Qc dn 6
(8) Rc DS 7,5
(9) (∃w)(Rw) EG 8
5
5. Proof with just UI and EG
Prove that the following argument is valid.
Hint: I think of using CP to prove some implications which will allow
the conclusion to be drawn by constructive dilemma.
1 (x)(P x ∨ Qx)
2 P a ∨ Qb
3 (v)(P v → Rv)
Goal: (∃y)(Qy) ∨ (∃z)(Rz)
Goal 1: P a → (∃z.Rz)
Goal 2: Qb → (∃y.Qy)
Notice that if I can prove these goals, the main goal follows by
constructive dilemma
Assume (4) P a for CP
(5) P a → Ra UI 3
(6) Ra MP 4,5
(7) (∃z)(Rz)
(8) P a → (∃z.Rz) CP 4-7
Assume (9) Qb for CP
(10) (∃y.Qy) EG 5
(11) Qb → (∃y.Qy) 9-10
(12) (∃y)(Qy) ∨ (∃z)(Rz) constructive dilemma 2,8,11
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6. Proof with EI
1 (x)(P x → Qx)
2 (∃z)(¬Qz)
3 (w)(P w ∨ Rw)
Goal: (∃x)(Rx)
(4) ¬Qa EI 2
(5) P a → Qa UI 1
(6) ¬P a MT 4,5
(7) P a ∨ Ra UI 3
(8) Ra DS 6,7
(9) (∃x)(Rx) EG 8
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7. Proof with UG
1 (x)(y)(P x → Qy)
2 Pa
3 (z)(Rz ∨ ¬Qz)
Goal: (w)(Rw)
(4) (y)(P a → Qy) UI 1
(5) P a → Qb UI 4
(6) Rb ∨ ¬Qb UI 3
(7) Qb MP 2,5
(8) ¬¬Qb dn 7
(9) Rb DS 8,6
(10) (w)(Rw) UG 9
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8. Proof with non-quantified statements with quantifier components.
Show that the argument is valid.
1 (x)(P x) → (x)(Qx)
2 Pa
3 (x)(P x → (w)(Rw))
4 (y)(Qy ∨ ¬Ry)
Goal: Ra
(5) P a → (w)(Rw) UI 3
(6) (w)(Rw) MP 2,5
(7) Ra UI 6
The trick here is that one cannot apply UI to 1; line 4, notice, is never
used.
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