Solution Set 3 - NYU

Solution Set 3 Problem 1 A. ∀X,Y s(X, Y ) ⇔ ∃Z c(X, Z) ∧ c(Y, Z). B. ∀X,Y g(X, Y ) ⇔ ∃Z c(X, Z) ∧ c(Z, Y ). C. ∀X ∃Y c(X, Y ). D. ∀X,Y s(X, Y ) ⇒ ∃Z g(X, Z) ∧ g(Y, Z). E. ∀X r(X) ⇒ c(X,e). F. ∀X,Y r(X) ∧ r(Y ) ⇒ s(X, Y ) Problem 2 The clauses generated from Skolemizing A, B, C and the negation of D are A.1. ¬s(X, Y ) ∨ c(X,sk0(X, Y )). A.2. ¬s(X, Y ) ∨ c(Y ,sk0(X, Y )). A.3. ¬c(X, Z) ∨ ¬c(Y, Z) ∨ s(X, Y ). B.1. ¬g(X, Y ) ∨ c(X,sk1(X, Y )). B.2. ¬g(X, Y ) ∨ c(sk1(X, Y ),Y ). B.3. ¬c(X, Z) ∨ ¬c(Z, Y ) ∨ g(X, Y ). C. c(X,sk2(X)). NotD.1 s(sk3,sk4). NotD.2 ¬g(sk3,Z) ∨ ¬g(sk4,Z). Resolution proof. G. c(sk3,sk0(sk3,sk4)) (From A.1 with NotD.1). H. c(sk4,sk0(sk3,sk4)). (From A.2 with NotD.1). I. ¬c(sk0(sk3,sk4),Y ) ∨ g(sk3,Y ) (From G with B.3 under substitution X→sk3, Z→sk0(sk3,sk4)).) J. g(sk3,sk2(sk0(sk3,sk4))). From I with C under substitution X→sk3, Y →sk2(sk0(sk3,sk4)).) K. ¬c(sk0(sk3,sk4),Y ) ∨ g(sk4,Y ) (From H with B.3 under substitution X→sk4, Z→sk0(sk3,sk4)).) L. g(sk4,sk2(sk0(sk3,sk4))). From K with C under substitution X→sk4, Y →sk2(sk0(sk3,sk4)).) M. ¬g(sk4,sk2(sk0(sk3,sk4))). (From J with NotD.2) N. ∅. (From L with M.)

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Problem 3 The clauses generated from Skolemizing A, B, C and the negation of D are A.1. ¬s(X, Y ) ∨ c(X,sk0(X, Y )). A.2. ¬s(X, Y ) ∨ c(Y ,sk0(X, Y )). A.3. ¬c(X, Z) ∨ ¬c(Y, Z) ∨ s(X, Y ). E. ¬r(X) ∨ c(X,e). NotF.1. r(sk5). NotF.2. r(sk6). NotF.3. s(sk5,sk6). Resolution proof: P. c(sk5,e). (From NotF.1 and E.) Q. c(sk6,e). (From NotF.2 and E.) R. ¬c(Y ,e) ∨ s(sk5,Y ). (From P and A.3). S. s(sk5,sk6). (From Q and R). T. ∅. (From S and NotF.3)

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