/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Left Termination of the query pattern p(g) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 8 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) PiDP (7) PiDPToQDPProof [SOUND, 14 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 56 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) UsableRulesReductionPairsProof [EQUIVALENT, 3 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Clauses: p(.(X, [])). p(.(s(s(X)), .(Y, Xs))) :- ','(p(.(X, .(Y, Xs))), p(.(s(s(s(s(Y)))), Xs))). p(.(0, Xs)) :- p(Xs). Query: p(g) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: p_in_1: (b) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_g(.(X, [])) -> p_out_g(.(X, [])) p_in_g(.(s(s(X)), .(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(.(X, .(Y, Xs)))) p_in_g(.(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) U3_g(Xs, p_out_g(Xs)) -> p_out_g(.(0, Xs)) U1_g(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(.(s(s(s(s(Y)))), Xs))) U2_g(X, Y, Xs, p_out_g(.(s(s(s(s(Y)))), Xs))) -> p_out_g(.(s(s(X)), .(Y, Xs))) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) .(x1, x2) = .(x1, x2) [] = [] p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 0 = 0 U3_g(x1, x2) = U3_g(x2) U2_g(x1, x2, x3, x4) = U2_g(x4) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: p_in_g(.(X, [])) -> p_out_g(.(X, [])) p_in_g(.(s(s(X)), .(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(.(X, .(Y, Xs)))) p_in_g(.(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) U3_g(Xs, p_out_g(Xs)) -> p_out_g(.(0, Xs)) U1_g(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(.(s(s(s(s(Y)))), Xs))) U2_g(X, Y, Xs, p_out_g(.(s(s(s(s(Y)))), Xs))) -> p_out_g(.(s(s(X)), .(Y, Xs))) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) .(x1, x2) = .(x1, x2) [] = [] p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 0 = 0 U3_g(x1, x2) = U3_g(x2) U2_g(x1, x2, x3, x4) = U2_g(x4) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: P_IN_G(.(s(s(X)), .(Y, Xs))) -> U1_G(X, Y, Xs, p_in_g(.(X, .(Y, Xs)))) P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) P_IN_G(.(0, Xs)) -> U3_G(Xs, p_in_g(Xs)) P_IN_G(.(0, Xs)) -> P_IN_G(Xs) U1_G(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> U2_G(X, Y, Xs, p_in_g(.(s(s(s(s(Y)))), Xs))) U1_G(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> P_IN_G(.(s(s(s(s(Y)))), Xs)) The TRS R consists of the following rules: p_in_g(.(X, [])) -> p_out_g(.(X, [])) p_in_g(.(s(s(X)), .(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(.(X, .(Y, Xs)))) p_in_g(.(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) U3_g(Xs, p_out_g(Xs)) -> p_out_g(.(0, Xs)) U1_g(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(.(s(s(s(s(Y)))), Xs))) U2_g(X, Y, Xs, p_out_g(.(s(s(s(s(Y)))), Xs))) -> p_out_g(.(s(s(X)), .(Y, Xs))) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) .(x1, x2) = .(x1, x2) [] = [] p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 0 = 0 U3_g(x1, x2) = U3_g(x2) U2_g(x1, x2, x3, x4) = U2_g(x4) P_IN_G(x1) = P_IN_G(x1) U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) U3_G(x1, x2) = U3_G(x2) U2_G(x1, x2, x3, x4) = U2_G(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_G(.(s(s(X)), .(Y, Xs))) -> U1_G(X, Y, Xs, p_in_g(.(X, .(Y, Xs)))) P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) P_IN_G(.(0, Xs)) -> U3_G(Xs, p_in_g(Xs)) P_IN_G(.(0, Xs)) -> P_IN_G(Xs) U1_G(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> U2_G(X, Y, Xs, p_in_g(.(s(s(s(s(Y)))), Xs))) U1_G(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> P_IN_G(.(s(s(s(s(Y)))), Xs)) The TRS R consists of the following rules: p_in_g(.(X, [])) -> p_out_g(.(X, [])) p_in_g(.(s(s(X)), .(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(.(X, .(Y, Xs)))) p_in_g(.(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) U3_g(Xs, p_out_g(Xs)) -> p_out_g(.(0, Xs)) U1_g(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(.(s(s(s(s(Y)))), Xs))) U2_g(X, Y, Xs, p_out_g(.(s(s(s(s(Y)))), Xs))) -> p_out_g(.(s(s(X)), .(Y, Xs))) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) .(x1, x2) = .(x1, x2) [] = [] p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 0 = 0 U3_g(x1, x2) = U3_g(x2) U2_g(x1, x2, x3, x4) = U2_g(x4) P_IN_G(x1) = P_IN_G(x1) U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) U3_G(x1, x2) = U3_G(x2) U2_G(x1, x2, x3, x4) = U2_G(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes. ---------------------------------------- (6) Obligation: Pi DP problem: The TRS P consists of the following rules: U1_G(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> P_IN_G(.(s(s(s(s(Y)))), Xs)) P_IN_G(.(s(s(X)), .(Y, Xs))) -> U1_G(X, Y, Xs, p_in_g(.(X, .(Y, Xs)))) P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) P_IN_G(.(0, Xs)) -> P_IN_G(Xs) The TRS R consists of the following rules: p_in_g(.(X, [])) -> p_out_g(.(X, [])) p_in_g(.(s(s(X)), .(Y, Xs))) -> U1_g(X, Y, Xs, p_in_g(.(X, .(Y, Xs)))) p_in_g(.(0, Xs)) -> U3_g(Xs, p_in_g(Xs)) U3_g(Xs, p_out_g(Xs)) -> p_out_g(.(0, Xs)) U1_g(X, Y, Xs, p_out_g(.(X, .(Y, Xs)))) -> U2_g(X, Y, Xs, p_in_g(.(s(s(s(s(Y)))), Xs))) U2_g(X, Y, Xs, p_out_g(.(s(s(s(s(Y)))), Xs))) -> p_out_g(.(s(s(X)), .(Y, Xs))) The argument filtering Pi contains the following mapping: p_in_g(x1) = p_in_g(x1) .(x1, x2) = .(x1, x2) [] = [] p_out_g(x1) = p_out_g s(x1) = s(x1) U1_g(x1, x2, x3, x4) = U1_g(x2, x3, x4) 0 = 0 U3_g(x1, x2) = U3_g(x2) U2_g(x1, x2, x3, x4) = U2_g(x4) P_IN_G(x1) = P_IN_G(x1) U1_G(x1, x2, x3, x4) = U1_G(x2, x3, x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: U1_G(Y, Xs, p_out_g) -> P_IN_G(.(s(s(s(s(Y)))), Xs)) P_IN_G(.(s(s(X)), .(Y, Xs))) -> U1_G(Y, Xs, p_in_g(.(X, .(Y, Xs)))) P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) P_IN_G(.(0, Xs)) -> P_IN_G(Xs) The TRS R consists of the following rules: p_in_g(.(X, [])) -> p_out_g p_in_g(.(s(s(X)), .(Y, Xs))) -> U1_g(Y, Xs, p_in_g(.(X, .(Y, Xs)))) p_in_g(.(0, Xs)) -> U3_g(p_in_g(Xs)) U3_g(p_out_g) -> p_out_g U1_g(Y, Xs, p_out_g) -> U2_g(p_in_g(.(s(s(s(s(Y)))), Xs))) U2_g(p_out_g) -> p_out_g The set Q consists of the following terms: p_in_g(x0) U3_g(x0) U1_g(x0, x1, x2) U2_g(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (9) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. P_IN_G(.(s(s(X)), .(Y, Xs))) -> U1_G(Y, Xs, p_in_g(.(X, .(Y, Xs)))) P_IN_G(.(0, Xs)) -> P_IN_G(Xs) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( U1_G_3(x_1, ..., x_3) ) = 2x_2 + 2 POL( p_in_g_1(x_1) ) = 0 POL( ._2(x_1, x_2) ) = x_2 + 2 POL( s_1(x_1) ) = 0 POL( U1_g_3(x_1, ..., x_3) ) = max{0, 2x_2 - 2} POL( 0 ) = 2 POL( U3_g_1(x_1) ) = 0 POL( U2_g_1(x_1) ) = 2x_1 POL( [] ) = 0 POL( p_out_g ) = 1 POL( P_IN_G_1(x_1) ) = max{0, 2x_1 - 2} The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: none ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: U1_G(Y, Xs, p_out_g) -> P_IN_G(.(s(s(s(s(Y)))), Xs)) P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) The TRS R consists of the following rules: p_in_g(.(X, [])) -> p_out_g p_in_g(.(s(s(X)), .(Y, Xs))) -> U1_g(Y, Xs, p_in_g(.(X, .(Y, Xs)))) p_in_g(.(0, Xs)) -> U3_g(p_in_g(Xs)) U3_g(p_out_g) -> p_out_g U1_g(Y, Xs, p_out_g) -> U2_g(p_in_g(.(s(s(s(s(Y)))), Xs))) U2_g(p_out_g) -> p_out_g The set Q consists of the following terms: p_in_g(x0) U3_g(x0) U1_g(x0, x1, x2) U2_g(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (11) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) The TRS R consists of the following rules: p_in_g(.(X, [])) -> p_out_g p_in_g(.(s(s(X)), .(Y, Xs))) -> U1_g(Y, Xs, p_in_g(.(X, .(Y, Xs)))) p_in_g(.(0, Xs)) -> U3_g(p_in_g(Xs)) U3_g(p_out_g) -> p_out_g U1_g(Y, Xs, p_out_g) -> U2_g(p_in_g(.(s(s(s(s(Y)))), Xs))) U2_g(p_out_g) -> p_out_g The set Q consists of the following terms: p_in_g(x0) U3_g(x0) U1_g(x0, x1, x2) U2_g(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (13) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) R is empty. The set Q consists of the following terms: p_in_g(x0) U3_g(x0) U1_g(x0, x1, x2) U2_g(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (15) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. p_in_g(x0) U3_g(x0) U1_g(x0, x1, x2) U2_g(x0) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (17) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. The following dependency pairs can be deleted: P_IN_G(.(s(s(X)), .(Y, Xs))) -> P_IN_G(.(X, .(Y, Xs))) No rules are removed from R. Used ordering: POLO with Polynomial interpretation [POLO]: POL(.(x_1, x_2)) = 2*x_1 + x_2 POL(P_IN_G(x_1)) = x_1 POL(s(x_1)) = 2*x_1 ---------------------------------------- (18) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (19) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (20) YES