/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.pl # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern perm(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 0 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 3 ms] (4) PiDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) PiDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) PiDP (10) PiDPToQDPProof [SOUND, 0 ms] (11) QDP (12) UsableRulesReductionPairsProof [EQUIVALENT, 2 ms] (13) QDP (14) PisEmptyProof [EQUIVALENT, 0 ms] (15) YES (16) PiDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) PiDP (19) PiDPToQDPProof [SOUND, 0 ms] (20) QDP (21) UsableRulesReductionPairsProof [EQUIVALENT, 4 ms] (22) QDP (23) PisEmptyProof [EQUIVALENT, 0 ms] (24) YES (25) PiDP (26) UsableRulesProof [EQUIVALENT, 0 ms] (27) PiDP (28) PiDPToQDPProof [SOUND, 0 ms] (29) QDP (30) MRRProof [EQUIVALENT, 0 ms] (31) QDP (32) PisEmptyProof [EQUIVALENT, 0 ms] (33) YES ---------------------------------------- (0) Obligation: Clauses: append2(parts([], Y), is(sum(Y))). append2(parts(.(H, X), Y), is(sum(.(H, Z)))) :- append2(parts(X, Y), is(sum(Z))). append1(parts([], Y), is(sum(Y))). append1(parts(.(H, X), Y), is(sum(.(H, Z)))) :- append1(parts(X, Y), is(sum(Z))). perm([], []). perm(L, .(H, T)) :- ','(append2(parts(V, .(H, U)), is(sum(L))), ','(append1(parts(V, U), is(sum(W))), perm(W, T))). Query: perm(g,a) ---------------------------------------- (1) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: perm_in_2: (b,f) append2_in_2: (f,b) append1_in_2: (b,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: perm_in_ga([], []) -> perm_out_ga([], []) perm_in_ga(L, .(H, T)) -> U3_ga(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) append2_in_ag(parts([], Y), is(sum(Y))) -> append2_out_ag(parts([], Y), is(sum(Y))) append2_in_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_ag(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) U1_ag(H, X, Y, Z, append2_out_ag(parts(X, Y), is(sum(Z)))) -> append2_out_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) U3_ga(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_ga(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) append1_in_ga(parts([], Y), is(sum(Y))) -> append1_out_ga(parts([], Y), is(sum(Y))) append1_in_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_ga(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) U2_ga(H, X, Y, Z, append1_out_ga(parts(X, Y), is(sum(Z)))) -> append1_out_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) U4_ga(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_ga(L, H, T, perm_in_ga(W, T)) U5_ga(L, H, T, perm_out_ga(W, T)) -> perm_out_ga(L, .(H, T)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) [] = [] perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append2_in_ag(x1, x2) = append2_in_ag(x2) .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) append2_out_ag(x1, x2) = append2_out_ag(x1) U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) parts(x1, x2) = parts(x1, x2) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) append1_in_ga(x1, x2) = append1_in_ga(x1) append1_out_ga(x1, x2) = append1_out_ga(x2) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(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: perm_in_ga([], []) -> perm_out_ga([], []) perm_in_ga(L, .(H, T)) -> U3_ga(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) append2_in_ag(parts([], Y), is(sum(Y))) -> append2_out_ag(parts([], Y), is(sum(Y))) append2_in_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_ag(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) U1_ag(H, X, Y, Z, append2_out_ag(parts(X, Y), is(sum(Z)))) -> append2_out_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) U3_ga(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_ga(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) append1_in_ga(parts([], Y), is(sum(Y))) -> append1_out_ga(parts([], Y), is(sum(Y))) append1_in_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_ga(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) U2_ga(H, X, Y, Z, append1_out_ga(parts(X, Y), is(sum(Z)))) -> append1_out_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) U4_ga(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_ga(L, H, T, perm_in_ga(W, T)) U5_ga(L, H, T, perm_out_ga(W, T)) -> perm_out_ga(L, .(H, T)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) [] = [] perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append2_in_ag(x1, x2) = append2_in_ag(x2) .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) append2_out_ag(x1, x2) = append2_out_ag(x1) U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) parts(x1, x2) = parts(x1, x2) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) append1_in_ga(x1, x2) = append1_in_ga(x1) append1_out_ga(x1, x2) = append1_out_ga(x2) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(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: PERM_IN_GA(L, .(H, T)) -> U3_GA(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) PERM_IN_GA(L, .(H, T)) -> APPEND2_IN_AG(parts(V, .(H, U)), is(sum(L))) APPEND2_IN_AG(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_AG(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) APPEND2_IN_AG(parts(.(H, X), Y), is(sum(.(H, Z)))) -> APPEND2_IN_AG(parts(X, Y), is(sum(Z))) U3_GA(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_GA(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) U3_GA(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> APPEND1_IN_GA(parts(V, U), is(sum(W))) APPEND1_IN_GA(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_GA(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) APPEND1_IN_GA(parts(.(H, X), Y), is(sum(.(H, Z)))) -> APPEND1_IN_GA(parts(X, Y), is(sum(Z))) U4_GA(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_GA(L, H, T, perm_in_ga(W, T)) U4_GA(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> PERM_IN_GA(W, T) The TRS R consists of the following rules: perm_in_ga([], []) -> perm_out_ga([], []) perm_in_ga(L, .(H, T)) -> U3_ga(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) append2_in_ag(parts([], Y), is(sum(Y))) -> append2_out_ag(parts([], Y), is(sum(Y))) append2_in_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_ag(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) U1_ag(H, X, Y, Z, append2_out_ag(parts(X, Y), is(sum(Z)))) -> append2_out_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) U3_ga(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_ga(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) append1_in_ga(parts([], Y), is(sum(Y))) -> append1_out_ga(parts([], Y), is(sum(Y))) append1_in_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_ga(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) U2_ga(H, X, Y, Z, append1_out_ga(parts(X, Y), is(sum(Z)))) -> append1_out_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) U4_ga(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_ga(L, H, T, perm_in_ga(W, T)) U5_ga(L, H, T, perm_out_ga(W, T)) -> perm_out_ga(L, .(H, T)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) [] = [] perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append2_in_ag(x1, x2) = append2_in_ag(x2) .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) append2_out_ag(x1, x2) = append2_out_ag(x1) U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) parts(x1, x2) = parts(x1, x2) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) append1_in_ga(x1, x2) = append1_in_ga(x1) append1_out_ga(x1, x2) = append1_out_ga(x2) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) APPEND2_IN_AG(x1, x2) = APPEND2_IN_AG(x2) U1_AG(x1, x2, x3, x4, x5) = U1_AG(x5) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) APPEND1_IN_GA(x1, x2) = APPEND1_IN_GA(x1) U2_GA(x1, x2, x3, x4, x5) = U2_GA(x5) U5_GA(x1, x2, x3, x4) = U5_GA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: PERM_IN_GA(L, .(H, T)) -> U3_GA(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) PERM_IN_GA(L, .(H, T)) -> APPEND2_IN_AG(parts(V, .(H, U)), is(sum(L))) APPEND2_IN_AG(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_AG(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) APPEND2_IN_AG(parts(.(H, X), Y), is(sum(.(H, Z)))) -> APPEND2_IN_AG(parts(X, Y), is(sum(Z))) U3_GA(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_GA(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) U3_GA(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> APPEND1_IN_GA(parts(V, U), is(sum(W))) APPEND1_IN_GA(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_GA(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) APPEND1_IN_GA(parts(.(H, X), Y), is(sum(.(H, Z)))) -> APPEND1_IN_GA(parts(X, Y), is(sum(Z))) U4_GA(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_GA(L, H, T, perm_in_ga(W, T)) U4_GA(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> PERM_IN_GA(W, T) The TRS R consists of the following rules: perm_in_ga([], []) -> perm_out_ga([], []) perm_in_ga(L, .(H, T)) -> U3_ga(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) append2_in_ag(parts([], Y), is(sum(Y))) -> append2_out_ag(parts([], Y), is(sum(Y))) append2_in_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_ag(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) U1_ag(H, X, Y, Z, append2_out_ag(parts(X, Y), is(sum(Z)))) -> append2_out_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) U3_ga(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_ga(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) append1_in_ga(parts([], Y), is(sum(Y))) -> append1_out_ga(parts([], Y), is(sum(Y))) append1_in_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_ga(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) U2_ga(H, X, Y, Z, append1_out_ga(parts(X, Y), is(sum(Z)))) -> append1_out_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) U4_ga(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_ga(L, H, T, perm_in_ga(W, T)) U5_ga(L, H, T, perm_out_ga(W, T)) -> perm_out_ga(L, .(H, T)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) [] = [] perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append2_in_ag(x1, x2) = append2_in_ag(x2) .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) append2_out_ag(x1, x2) = append2_out_ag(x1) U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) parts(x1, x2) = parts(x1, x2) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) append1_in_ga(x1, x2) = append1_in_ga(x1) append1_out_ga(x1, x2) = append1_out_ga(x2) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) APPEND2_IN_AG(x1, x2) = APPEND2_IN_AG(x2) U1_AG(x1, x2, x3, x4, x5) = U1_AG(x5) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) APPEND1_IN_GA(x1, x2) = APPEND1_IN_GA(x1) U2_GA(x1, x2, x3, x4, x5) = U2_GA(x5) U5_GA(x1, x2, x3, x4) = U5_GA(x4) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 5 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND1_IN_GA(parts(.(H, X), Y), is(sum(.(H, Z)))) -> APPEND1_IN_GA(parts(X, Y), is(sum(Z))) The TRS R consists of the following rules: perm_in_ga([], []) -> perm_out_ga([], []) perm_in_ga(L, .(H, T)) -> U3_ga(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) append2_in_ag(parts([], Y), is(sum(Y))) -> append2_out_ag(parts([], Y), is(sum(Y))) append2_in_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_ag(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) U1_ag(H, X, Y, Z, append2_out_ag(parts(X, Y), is(sum(Z)))) -> append2_out_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) U3_ga(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_ga(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) append1_in_ga(parts([], Y), is(sum(Y))) -> append1_out_ga(parts([], Y), is(sum(Y))) append1_in_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_ga(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) U2_ga(H, X, Y, Z, append1_out_ga(parts(X, Y), is(sum(Z)))) -> append1_out_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) U4_ga(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_ga(L, H, T, perm_in_ga(W, T)) U5_ga(L, H, T, perm_out_ga(W, T)) -> perm_out_ga(L, .(H, T)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) [] = [] perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append2_in_ag(x1, x2) = append2_in_ag(x2) .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) append2_out_ag(x1, x2) = append2_out_ag(x1) U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) parts(x1, x2) = parts(x1, x2) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) append1_in_ga(x1, x2) = append1_in_ga(x1) append1_out_ga(x1, x2) = append1_out_ga(x2) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) APPEND1_IN_GA(x1, x2) = APPEND1_IN_GA(x1) We have to consider all (P,R,Pi)-chains ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (9) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND1_IN_GA(parts(.(H, X), Y), is(sum(.(H, Z)))) -> APPEND1_IN_GA(parts(X, Y), is(sum(Z))) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) parts(x1, x2) = parts(x1, x2) APPEND1_IN_GA(x1, x2) = APPEND1_IN_GA(x1) We have to consider all (P,R,Pi)-chains ---------------------------------------- (10) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: APPEND1_IN_GA(parts(.(X), Y)) -> APPEND1_IN_GA(parts(X, Y)) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (12) 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: APPEND1_IN_GA(parts(.(X), Y)) -> APPEND1_IN_GA(parts(X, Y)) No rules are removed from R. Used ordering: POLO with Polynomial interpretation [POLO]: POL(.(x_1)) = x_1 POL(APPEND1_IN_GA(x_1)) = 2*x_1 POL(parts(x_1, x_2)) = 2*x_1 + x_2 ---------------------------------------- (13) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (14) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND2_IN_AG(parts(.(H, X), Y), is(sum(.(H, Z)))) -> APPEND2_IN_AG(parts(X, Y), is(sum(Z))) The TRS R consists of the following rules: perm_in_ga([], []) -> perm_out_ga([], []) perm_in_ga(L, .(H, T)) -> U3_ga(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) append2_in_ag(parts([], Y), is(sum(Y))) -> append2_out_ag(parts([], Y), is(sum(Y))) append2_in_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_ag(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) U1_ag(H, X, Y, Z, append2_out_ag(parts(X, Y), is(sum(Z)))) -> append2_out_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) U3_ga(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_ga(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) append1_in_ga(parts([], Y), is(sum(Y))) -> append1_out_ga(parts([], Y), is(sum(Y))) append1_in_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_ga(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) U2_ga(H, X, Y, Z, append1_out_ga(parts(X, Y), is(sum(Z)))) -> append1_out_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) U4_ga(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_ga(L, H, T, perm_in_ga(W, T)) U5_ga(L, H, T, perm_out_ga(W, T)) -> perm_out_ga(L, .(H, T)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) [] = [] perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append2_in_ag(x1, x2) = append2_in_ag(x2) .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) append2_out_ag(x1, x2) = append2_out_ag(x1) U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) parts(x1, x2) = parts(x1, x2) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) append1_in_ga(x1, x2) = append1_in_ga(x1) append1_out_ga(x1, x2) = append1_out_ga(x2) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) APPEND2_IN_AG(x1, x2) = APPEND2_IN_AG(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (17) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (18) Obligation: Pi DP problem: The TRS P consists of the following rules: APPEND2_IN_AG(parts(.(H, X), Y), is(sum(.(H, Z)))) -> APPEND2_IN_AG(parts(X, Y), is(sum(Z))) R is empty. The argument filtering Pi contains the following mapping: .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) parts(x1, x2) = parts(x1, x2) APPEND2_IN_AG(x1, x2) = APPEND2_IN_AG(x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (19) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: APPEND2_IN_AG(is(sum(.(Z)))) -> APPEND2_IN_AG(is(sum(Z))) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (21) 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: APPEND2_IN_AG(is(sum(.(Z)))) -> APPEND2_IN_AG(is(sum(Z))) No rules are removed from R. Used ordering: POLO with Polynomial interpretation [POLO]: POL(.(x_1)) = 2*x_1 POL(APPEND2_IN_AG(x_1)) = 2*x_1 POL(is(x_1)) = x_1 POL(sum(x_1)) = x_1 ---------------------------------------- (22) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (23) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (24) YES ---------------------------------------- (25) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_GA(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_GA(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) U4_GA(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> PERM_IN_GA(W, T) PERM_IN_GA(L, .(H, T)) -> U3_GA(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) The TRS R consists of the following rules: perm_in_ga([], []) -> perm_out_ga([], []) perm_in_ga(L, .(H, T)) -> U3_ga(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) append2_in_ag(parts([], Y), is(sum(Y))) -> append2_out_ag(parts([], Y), is(sum(Y))) append2_in_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_ag(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) U1_ag(H, X, Y, Z, append2_out_ag(parts(X, Y), is(sum(Z)))) -> append2_out_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) U3_ga(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_ga(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) append1_in_ga(parts([], Y), is(sum(Y))) -> append1_out_ga(parts([], Y), is(sum(Y))) append1_in_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_ga(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) U2_ga(H, X, Y, Z, append1_out_ga(parts(X, Y), is(sum(Z)))) -> append1_out_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) U4_ga(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> U5_ga(L, H, T, perm_in_ga(W, T)) U5_ga(L, H, T, perm_out_ga(W, T)) -> perm_out_ga(L, .(H, T)) The argument filtering Pi contains the following mapping: perm_in_ga(x1, x2) = perm_in_ga(x1) [] = [] perm_out_ga(x1, x2) = perm_out_ga(x2) U3_ga(x1, x2, x3, x4) = U3_ga(x4) append2_in_ag(x1, x2) = append2_in_ag(x2) .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) append2_out_ag(x1, x2) = append2_out_ag(x1) U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) parts(x1, x2) = parts(x1, x2) U4_ga(x1, x2, x3, x4, x5, x6) = U4_ga(x6) append1_in_ga(x1, x2) = append1_in_ga(x1) append1_out_ga(x1, x2) = append1_out_ga(x2) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) U5_ga(x1, x2, x3, x4) = U5_ga(x4) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) We have to consider all (P,R,Pi)-chains ---------------------------------------- (26) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (27) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_GA(L, H, T, append2_out_ag(parts(V, .(H, U)), is(sum(L)))) -> U4_GA(L, H, T, V, U, append1_in_ga(parts(V, U), is(sum(W)))) U4_GA(L, H, T, V, U, append1_out_ga(parts(V, U), is(sum(W)))) -> PERM_IN_GA(W, T) PERM_IN_GA(L, .(H, T)) -> U3_GA(L, H, T, append2_in_ag(parts(V, .(H, U)), is(sum(L)))) The TRS R consists of the following rules: append1_in_ga(parts([], Y), is(sum(Y))) -> append1_out_ga(parts([], Y), is(sum(Y))) append1_in_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U2_ga(H, X, Y, Z, append1_in_ga(parts(X, Y), is(sum(Z)))) append2_in_ag(parts([], Y), is(sum(Y))) -> append2_out_ag(parts([], Y), is(sum(Y))) append2_in_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) -> U1_ag(H, X, Y, Z, append2_in_ag(parts(X, Y), is(sum(Z)))) U2_ga(H, X, Y, Z, append1_out_ga(parts(X, Y), is(sum(Z)))) -> append1_out_ga(parts(.(H, X), Y), is(sum(.(H, Z)))) U1_ag(H, X, Y, Z, append2_out_ag(parts(X, Y), is(sum(Z)))) -> append2_out_ag(parts(.(H, X), Y), is(sum(.(H, Z)))) The argument filtering Pi contains the following mapping: [] = [] append2_in_ag(x1, x2) = append2_in_ag(x2) .(x1, x2) = .(x2) is(x1) = is(x1) sum(x1) = sum(x1) append2_out_ag(x1, x2) = append2_out_ag(x1) U1_ag(x1, x2, x3, x4, x5) = U1_ag(x5) parts(x1, x2) = parts(x1, x2) append1_in_ga(x1, x2) = append1_in_ga(x1) append1_out_ga(x1, x2) = append1_out_ga(x2) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) PERM_IN_GA(x1, x2) = PERM_IN_GA(x1) U3_GA(x1, x2, x3, x4) = U3_GA(x4) U4_GA(x1, x2, x3, x4, x5, x6) = U4_GA(x6) We have to consider all (P,R,Pi)-chains ---------------------------------------- (28) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (29) Obligation: Q DP problem: The TRS P consists of the following rules: U3_GA(append2_out_ag(parts(V, .(U)))) -> U4_GA(append1_in_ga(parts(V, U))) U4_GA(append1_out_ga(is(sum(W)))) -> PERM_IN_GA(W) PERM_IN_GA(L) -> U3_GA(append2_in_ag(is(sum(L)))) The TRS R consists of the following rules: append1_in_ga(parts([], Y)) -> append1_out_ga(is(sum(Y))) append1_in_ga(parts(.(X), Y)) -> U2_ga(append1_in_ga(parts(X, Y))) append2_in_ag(is(sum(Y))) -> append2_out_ag(parts([], Y)) append2_in_ag(is(sum(.(Z)))) -> U1_ag(append2_in_ag(is(sum(Z)))) U2_ga(append1_out_ga(is(sum(Z)))) -> append1_out_ga(is(sum(.(Z)))) U1_ag(append2_out_ag(parts(X, Y))) -> append2_out_ag(parts(.(X), Y)) The set Q consists of the following terms: append1_in_ga(x0) append2_in_ag(x0) U2_ga(x0) U1_ag(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (30) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: U3_GA(append2_out_ag(parts(V, .(U)))) -> U4_GA(append1_in_ga(parts(V, U))) U4_GA(append1_out_ga(is(sum(W)))) -> PERM_IN_GA(W) PERM_IN_GA(L) -> U3_GA(append2_in_ag(is(sum(L)))) Strictly oriented rules of the TRS R: append1_in_ga(parts([], Y)) -> append1_out_ga(is(sum(Y))) append1_in_ga(parts(.(X), Y)) -> U2_ga(append1_in_ga(parts(X, Y))) append2_in_ag(is(sum(Y))) -> append2_out_ag(parts([], Y)) append2_in_ag(is(sum(.(Z)))) -> U1_ag(append2_in_ag(is(sum(Z)))) U2_ga(append1_out_ga(is(sum(Z)))) -> append1_out_ga(is(sum(.(Z)))) U1_ag(append2_out_ag(parts(X, Y))) -> append2_out_ag(parts(.(X), Y)) Used ordering: Knuth-Bendix order [KBO] with precedence:parts_2 > append1_in_ga_1 > U2_ga_1 > PERM_IN_GA_1 > ._1 > is_1 > append1_out_ga_1 > append2_in_ag_1 > U1_ag_1 > append2_out_ag_1 > U3_GA_1 > U4_GA_1 > sum_1 > [] and weight map: []=15 append1_in_ga_1=8 append1_out_ga_1=5 is_1=7 sum_1=7 ._1=6 U2_ga_1=6 append2_in_ag_1=2 append2_out_ag_1=1 U1_ag_1=6 U3_GA_1=2 U4_GA_1=1 PERM_IN_GA_1=19 parts_2=0 The variable weight is 5 ---------------------------------------- (31) Obligation: Q DP problem: P is empty. R is empty. The set Q consists of the following terms: append1_in_ga(x0) append2_in_ag(x0) U2_ga(x0) U1_ag(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (32) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (33) YES