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 4.(1)(A→(B→C) (3)A (4)(B→C) (1),(3)MP (5)C (2)、(4)MP What we have demonstrated is (A-(B→C,BA)上C so by the Deduction Theorem, We have (A→(B→C)→(B→(A→C)) (5 pts.) 5. We can define an interpretation I as Follows D=Z, A(x)stands for x<0, and f,(x)stands for the successor of x (Spts.) So the wf.(V x1)(A (x)A' (f (xD)has the interpretation for all xe Di, if x<o then the successor of x<o Obviously, when x=-I this is false. (5 pts. =.Proof.(each 10 points, total 30 points) 1. Proof. Now suppose, conversely,that(A1…∧An)→A) is a tautology and that Al,An;∴A is not a valid argument form. Then there is an assignment of truth values which makes each Ai (1≤isn) take value t and makes a take value p, so that(A1∧.∧Aan)→A) takes value F a valid argument form. (5 pt 2. Proof. If B is a contradiction, then(B)is obviously tautology, i.e. H-B)(5pts. )Suppose that B is a theorem of any consistent extension of L, then B and(B) are both the theorems of L', But this result contradicts that L' is a consistent extension of L, so if b is a contradiction, then it cannot be a theorem of any consistent extension of L (5 pts 3. Proof. Let v be a valuation in I. Then v satisfies A and v satisfies(AB). By the definition o 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） 4. (1) (A→(B→C)) as. (2) B as. (3) A as. (4) (B→C) (1),(3)MP (5) C (2),(4)MP (5 pts.) What we have demonstrated is {(A→(B→C),B,A)}├C so by the Deduction Theorem, We have ((A→(B→C) →(B→(A→C))) (5 pts.) 5. We can define an interpretation I as Follows. DI = Z , A 1 1 (x) stands for x<0, and f 1 1 (x) stands for the successor of x. (5pts.) So the wf. (  x1) ( 1 A1 (x1)→ 1 A1 ( 1 1 f (x1))) has the interpretation for all x  DI, if x＜0 then the successor of x＜0. Obviously, when x = -1 this is false. (5 pts.) 三. Proof. (each 10 points, total 30 points) 1. Proof. Now suppose, conversely, that ((A1∧…∧An) → A) is a tautology and that A1,…,An;∴A is not a valid argument form. Then there is an assignment of truth values which makes each Ai (1≤i≤n) take value T and makes A take value F, so that ((A1∧…∧An)→A) takes value F. (5 pts.)This contradicts the assumption that this statement form is a tautology, and so A1,…, An; ∴A is a valid argument form. (5 pts.) 2. Proof. If B is a contradiction, then (～B)is obviously tautology, i.e.├(～B) (5.pts.) Suppose that B is a theorem of any consistent extension of L, then B and (～B) are both the theorems of L’, But this result contradicts that L’ is a consistent extension of L, so if B is a contradiction, then it cannot be a theorem of any consistent extension of L. (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