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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. 1 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 2 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) 4 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 6 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 7 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 8 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. 9