i.→D→-B. i.C→ iv, - A (b) The result is -A. The tableau proof is shown in Figure 3.(2 marks) F(=A) T A TA→(BVC) FA T(BVC) 6 TB TC TC→-D FC T-D Figure 3: Tableau proof As Figure 3 shown, there is at least a noncontradictory path. So the argument is not logically valid. (1 marks 3. Given a formula (, y, a)=(v)()va(a, y, a)),p(a, y, a), where a, 3, z are free variables. Answer the following questions (8m (a) Show which occurrence of every variable is free. If it is bound, show which quantifier bounds it (b)Given a function f(t, a) and t does not occurs in Is y/f(t, a) substitutable in p(a, y, a)? Show your reason. (2 marks (c) Discuss its truth of (a, y, 2). If it is not valid, give a model made it false. (3 marks) Solution.(a)The third occurrence of the second occurrence of z and both occurrences of y are free. (1 mark) The first two a is bounded by(Vr).(1 mark)The first z is bounded by 3z. (1 mark (b) The second substitution of y is substitutable for z, t in f is still free. However the first one is not for z is bounded by 3x(1 mark) Then it is not substitutable. (1 mark) (c) It is not a sentence. We first represent it as the V-closure form. It is valid. It could be proved by a tableau proof shown in Figure 4. Obviously, the tableau noncontradictory. (2 marks)So it is not valid. The counterexample is very simple Let A=co, C1, c21,v=(co1, p =0.(1 markii. ¬D → ¬B. iii. C → ¬D. iv. ¬A. (b) The result is ¬A. The tableau proof is shown in Figure 3.(2 marks) F (¬A) T A T A → (B ∨ C) F A T (B ∨ C) ⊗ T B T C . . . T C → ¬D F C T¬D ⊗ F D Figure 3: Tableau proof As Figure 3 shown, there is at least a noncontradictory path. So the argument is not logically valid.(1 marks) 3. Given a formula Φ(x, y, z) = (∀x)(ψ(x)∨(∃z)φ(x, y, z)) → φ(x, y, z), where x, y, z are free variables. Answer the following questions. (8 marks) (a) Show which occurrence of every variable is free. If it is bound, show which quantifier bounds it. (3 marks) (b) Given a function f(t, z) and t does not occurs in Φ. Is y/f(t, z) substitutable in Φ(x, y, z)? Show your reason. (2 marks) (c) Discuss its truth of Φ(x, y, z).If it is not valid, give a model made it false. (3 marks) Solution. (a) The third occurrence of x, the second occurrence of z and both occurrences of y are free.(1 mark) The first two x is bounded by (∀x). (1 mark) The first z is bounded by ∃z.(1 mark) (b) The second substitution of y is substitutable for z, t in f is still free. However the first one is not for z is bounded by ∃z.(1 mark) Then it is not substitutable. (1 mark) (c) It is not a sentence. We first represent it as the ∀-closure form. It is valid. It could be proved by a tableau proof shown in Figure 4. Obviously, the tableau is noncontradictory.(2 marks) So it is not valid. The counterexample is very simple. Let A = {c0, c1, c2}, ψ = {c0}, φ = ∅. (1 mark) 3