湖北大学2004—2005学年度第二学期课程考试试题参考谷案及评分标准 argument is valid. (5 pts. 4.(1)(A→B) (3)A (4)B (1),(3)MP (5)(~(B→C)→(~A)→(A→(B→C)(La (2),(SMP (8)C (4)(7)MP(5pts) What we have demonstrated (A→B(~(B→C)→~AA}C So by the deduction Theorem, we have (A→B)→((~(B→C)→~A)→(A→C)) 5. We can define an interpretation I as follows DI=N, and A,(x, y) Stands for x<y, (5 pts. So the wf.(v Xi)(AF(x1, X2)-AF(x2, xi) has the interpretation for all x and yE D if x <y then Obviously, this is false (5pts) = Proof (each 10 points, total 30 points) 1. Proof. Suppose first that Al,…,An;∴ a is a valid argument form and that(A1∧…∧,An)→A is not a tautology, Then there is an assignment of truth values to the statement variables occurring such that(A1A., )takes value T, and A takes value F (5 pts. For this assignment, then, each Ai takes value T(lsis), and A takes value F. This contradicts the validity of the argument form and so(A1∧.∧An)→A) must be a tautology.(Spts.) 2. Proof. Suppose that L is not consistent, i.e. that there is a wf. A such that FA and +(A).Then by Proposition 2. 14(The Soundness Theorem), both A and(A)are tautologies. (5 pts. )This is impossible since if A is a tautology then(A) is a contradiction and if(A)is a tautology then a is a contradiction. Thus L must be consistent. (5 pts. 3. Proof. Let v be a valuation in I. Then v satisfies A and v satisfies(A-B). By the definition satisfaction, then, either v satisfies( A)or v satisfies B. But v cannot satisfy(A),so v must satisfy B (5 pts. It follows that is satisfied by every valuation in I, so B is true in I (5 pts.) 第2页共2页)湖北大学 2004——2005 学年度第二学期课程考试试题 参考答案及评分标准 （第 2 页 共 2 页） argument is valid. (5 pts.) 4. (1) (A→B) as. (2)～(B→C)→(～A) as. (3)A as. (4)B (1),(3)MP (5)(～(B→C)→(～A))→(A→(B→C)) (L3) (6)A→(B→C) (2),(5)MP (7)(B→C) (3),(6)MP (8)C (4)(7)MP (5pts.) What we have demonstrated is {(A→B),(～(B→C)→～A),A} ├ C So by the Deduction Theorem, we have ((A→B)→((～(B→C)→～A)→(A→C))) 5. We can define an interpretation I as follows. DI = N, and 2 A1 (x,y) Stands for x＜y, (5 pts.) So the wf. (  x1) ( 2 A1 (x1,x2)→ 2 A1 (x2,x1)) has the interpretation for all x and y  DI, if x＜y then y＜x. Obviously, this is false. (5pts.) 三. Proof. (each 10 points, total 30 points) 1. Proof. Suppose first that A1,…,An; ∴A is a valid argument form and that ((A1∧…∧,An)→A) is not a tautology, Then there is an assignment of truth values to the statement variables occurring such that (A1∧…∧An)takes value T, and A takes value F. (5 pts.) For this assignment, then, each Ai takes value T(1≤i≤n), and A takes value F. This contradicts the validity of the argument form, and so ((A1∧…∧An)→A) must be a tautology. (5 pts.) 2. Proof. Suppose that L is not consistent, i.e. that there is a wf. A such that ├ A and ├ (～A). Then by Proposition 2.14 (The Soundness Theorem), both A and (～A) are tautologies.(5 pts.) This is impossible since if A is a tautology then (～A) is a contradiction and if (～A) is a tautology then A is a contradiction. Thus L must be consistent. (5 pts.) 3. Proof. Let v be a valuation in I. Then v satisfies A and v satisfies (A→B). By the definition of satisfaction, then ,either v satisfies (～A)or v satisfies B. But v cannot satisfy (～A), so v must satisfy B. (5 pts.) It follows that is satisfied by every valuation in I, so B is true in I. (5 pts.) L L L