/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern factor(g,a) w.r.t. the given Prolog program could successfully be proven: (0) Prolog (1) PrologToPiTRSProof [SOUND, 23 ms] (2) PiTRS (3) DependencyPairsProof [EQUIVALENT, 0 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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) PiDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) PiDP (17) PiDPToQDPProof [SOUND, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES (21) PiDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) PiDP (24) PiDPToQDPProof [SOUND, 0 ms] (25) QDP (26) QDPOrderProof [EQUIVALENT, 24 ms] (27) QDP (28) PisEmptyProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Clauses: factor(cons(X, nil), X). factor(cons(X, cons(Y, Xs)), T) :- ','(times(X, Y, Z), factor(cons(Z, Xs), T)). times(0, X_, 0). times(s(X), Y, Z) :- ','(times(X, Y, XY), plus(XY, Y, Z)). plus(0, X, X). plus(s(X), Y, s(Z)) :- plus(X, Y, Z). Query: factor(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: factor_in_2: (b,f) times_in_3: (b,b,f) plus_in_3: (b,b,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: factor_in_ga(cons(X, nil), X) -> factor_out_ga(cons(X, nil), X) factor_in_ga(cons(X, cons(Y, Xs)), T) -> U1_ga(X, Y, Xs, T, times_in_gga(X, Y, Z)) times_in_gga(0, X_, 0) -> times_out_gga(0, X_, 0) times_in_gga(s(X), Y, Z) -> U3_gga(X, Y, Z, times_in_gga(X, Y, XY)) U3_gga(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_gga(X, Y, Z, plus_in_gga(XY, Y, Z)) plus_in_gga(0, X, X) -> plus_out_gga(0, X, X) plus_in_gga(s(X), Y, s(Z)) -> U5_gga(X, Y, Z, plus_in_gga(X, Y, Z)) U5_gga(X, Y, Z, plus_out_gga(X, Y, Z)) -> plus_out_gga(s(X), Y, s(Z)) U4_gga(X, Y, Z, plus_out_gga(XY, Y, Z)) -> times_out_gga(s(X), Y, Z) U1_ga(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_ga(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U2_ga(X, Y, Xs, T, factor_out_ga(cons(Z, Xs), T)) -> factor_out_ga(cons(X, cons(Y, Xs)), T) The argument filtering Pi contains the following mapping: factor_in_ga(x1, x2) = factor_in_ga(x1) cons(x1, x2) = cons(x1, x2) nil = nil factor_out_ga(x1, x2) = factor_out_ga(x2) U1_ga(x1, x2, x3, x4, x5) = U1_ga(x3, x5) times_in_gga(x1, x2, x3) = times_in_gga(x1, x2) 0 = 0 times_out_gga(x1, x2, x3) = times_out_gga(x3) s(x1) = s(x1) U3_gga(x1, x2, x3, x4) = U3_gga(x2, x4) U4_gga(x1, x2, x3, x4) = U4_gga(x4) plus_in_gga(x1, x2, x3) = plus_in_gga(x1, x2) plus_out_gga(x1, x2, x3) = plus_out_gga(x3) U5_gga(x1, x2, x3, x4) = U5_gga(x4) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (2) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: factor_in_ga(cons(X, nil), X) -> factor_out_ga(cons(X, nil), X) factor_in_ga(cons(X, cons(Y, Xs)), T) -> U1_ga(X, Y, Xs, T, times_in_gga(X, Y, Z)) times_in_gga(0, X_, 0) -> times_out_gga(0, X_, 0) times_in_gga(s(X), Y, Z) -> U3_gga(X, Y, Z, times_in_gga(X, Y, XY)) U3_gga(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_gga(X, Y, Z, plus_in_gga(XY, Y, Z)) plus_in_gga(0, X, X) -> plus_out_gga(0, X, X) plus_in_gga(s(X), Y, s(Z)) -> U5_gga(X, Y, Z, plus_in_gga(X, Y, Z)) U5_gga(X, Y, Z, plus_out_gga(X, Y, Z)) -> plus_out_gga(s(X), Y, s(Z)) U4_gga(X, Y, Z, plus_out_gga(XY, Y, Z)) -> times_out_gga(s(X), Y, Z) U1_ga(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_ga(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U2_ga(X, Y, Xs, T, factor_out_ga(cons(Z, Xs), T)) -> factor_out_ga(cons(X, cons(Y, Xs)), T) The argument filtering Pi contains the following mapping: factor_in_ga(x1, x2) = factor_in_ga(x1) cons(x1, x2) = cons(x1, x2) nil = nil factor_out_ga(x1, x2) = factor_out_ga(x2) U1_ga(x1, x2, x3, x4, x5) = U1_ga(x3, x5) times_in_gga(x1, x2, x3) = times_in_gga(x1, x2) 0 = 0 times_out_gga(x1, x2, x3) = times_out_gga(x3) s(x1) = s(x1) U3_gga(x1, x2, x3, x4) = U3_gga(x2, x4) U4_gga(x1, x2, x3, x4) = U4_gga(x4) plus_in_gga(x1, x2, x3) = plus_in_gga(x1, x2) plus_out_gga(x1, x2, x3) = plus_out_gga(x3) U5_gga(x1, x2, x3, x4) = U5_gga(x4) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) ---------------------------------------- (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: FACTOR_IN_GA(cons(X, cons(Y, Xs)), T) -> U1_GA(X, Y, Xs, T, times_in_gga(X, Y, Z)) FACTOR_IN_GA(cons(X, cons(Y, Xs)), T) -> TIMES_IN_GGA(X, Y, Z) TIMES_IN_GGA(s(X), Y, Z) -> U3_GGA(X, Y, Z, times_in_gga(X, Y, XY)) TIMES_IN_GGA(s(X), Y, Z) -> TIMES_IN_GGA(X, Y, XY) U3_GGA(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_GGA(X, Y, Z, plus_in_gga(XY, Y, Z)) U3_GGA(X, Y, Z, times_out_gga(X, Y, XY)) -> PLUS_IN_GGA(XY, Y, Z) PLUS_IN_GGA(s(X), Y, s(Z)) -> U5_GGA(X, Y, Z, plus_in_gga(X, Y, Z)) PLUS_IN_GGA(s(X), Y, s(Z)) -> PLUS_IN_GGA(X, Y, Z) U1_GA(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_GA(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U1_GA(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> FACTOR_IN_GA(cons(Z, Xs), T) The TRS R consists of the following rules: factor_in_ga(cons(X, nil), X) -> factor_out_ga(cons(X, nil), X) factor_in_ga(cons(X, cons(Y, Xs)), T) -> U1_ga(X, Y, Xs, T, times_in_gga(X, Y, Z)) times_in_gga(0, X_, 0) -> times_out_gga(0, X_, 0) times_in_gga(s(X), Y, Z) -> U3_gga(X, Y, Z, times_in_gga(X, Y, XY)) U3_gga(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_gga(X, Y, Z, plus_in_gga(XY, Y, Z)) plus_in_gga(0, X, X) -> plus_out_gga(0, X, X) plus_in_gga(s(X), Y, s(Z)) -> U5_gga(X, Y, Z, plus_in_gga(X, Y, Z)) U5_gga(X, Y, Z, plus_out_gga(X, Y, Z)) -> plus_out_gga(s(X), Y, s(Z)) U4_gga(X, Y, Z, plus_out_gga(XY, Y, Z)) -> times_out_gga(s(X), Y, Z) U1_ga(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_ga(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U2_ga(X, Y, Xs, T, factor_out_ga(cons(Z, Xs), T)) -> factor_out_ga(cons(X, cons(Y, Xs)), T) The argument filtering Pi contains the following mapping: factor_in_ga(x1, x2) = factor_in_ga(x1) cons(x1, x2) = cons(x1, x2) nil = nil factor_out_ga(x1, x2) = factor_out_ga(x2) U1_ga(x1, x2, x3, x4, x5) = U1_ga(x3, x5) times_in_gga(x1, x2, x3) = times_in_gga(x1, x2) 0 = 0 times_out_gga(x1, x2, x3) = times_out_gga(x3) s(x1) = s(x1) U3_gga(x1, x2, x3, x4) = U3_gga(x2, x4) U4_gga(x1, x2, x3, x4) = U4_gga(x4) plus_in_gga(x1, x2, x3) = plus_in_gga(x1, x2) plus_out_gga(x1, x2, x3) = plus_out_gga(x3) U5_gga(x1, x2, x3, x4) = U5_gga(x4) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) FACTOR_IN_GA(x1, x2) = FACTOR_IN_GA(x1) U1_GA(x1, x2, x3, x4, x5) = U1_GA(x3, x5) TIMES_IN_GGA(x1, x2, x3) = TIMES_IN_GGA(x1, x2) U3_GGA(x1, x2, x3, x4) = U3_GGA(x2, x4) U4_GGA(x1, x2, x3, x4) = U4_GGA(x4) PLUS_IN_GGA(x1, x2, x3) = PLUS_IN_GGA(x1, x2) U5_GGA(x1, x2, x3, x4) = U5_GGA(x4) U2_GA(x1, x2, x3, x4, x5) = U2_GA(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (4) Obligation: Pi DP problem: The TRS P consists of the following rules: FACTOR_IN_GA(cons(X, cons(Y, Xs)), T) -> U1_GA(X, Y, Xs, T, times_in_gga(X, Y, Z)) FACTOR_IN_GA(cons(X, cons(Y, Xs)), T) -> TIMES_IN_GGA(X, Y, Z) TIMES_IN_GGA(s(X), Y, Z) -> U3_GGA(X, Y, Z, times_in_gga(X, Y, XY)) TIMES_IN_GGA(s(X), Y, Z) -> TIMES_IN_GGA(X, Y, XY) U3_GGA(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_GGA(X, Y, Z, plus_in_gga(XY, Y, Z)) U3_GGA(X, Y, Z, times_out_gga(X, Y, XY)) -> PLUS_IN_GGA(XY, Y, Z) PLUS_IN_GGA(s(X), Y, s(Z)) -> U5_GGA(X, Y, Z, plus_in_gga(X, Y, Z)) PLUS_IN_GGA(s(X), Y, s(Z)) -> PLUS_IN_GGA(X, Y, Z) U1_GA(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_GA(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U1_GA(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> FACTOR_IN_GA(cons(Z, Xs), T) The TRS R consists of the following rules: factor_in_ga(cons(X, nil), X) -> factor_out_ga(cons(X, nil), X) factor_in_ga(cons(X, cons(Y, Xs)), T) -> U1_ga(X, Y, Xs, T, times_in_gga(X, Y, Z)) times_in_gga(0, X_, 0) -> times_out_gga(0, X_, 0) times_in_gga(s(X), Y, Z) -> U3_gga(X, Y, Z, times_in_gga(X, Y, XY)) U3_gga(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_gga(X, Y, Z, plus_in_gga(XY, Y, Z)) plus_in_gga(0, X, X) -> plus_out_gga(0, X, X) plus_in_gga(s(X), Y, s(Z)) -> U5_gga(X, Y, Z, plus_in_gga(X, Y, Z)) U5_gga(X, Y, Z, plus_out_gga(X, Y, Z)) -> plus_out_gga(s(X), Y, s(Z)) U4_gga(X, Y, Z, plus_out_gga(XY, Y, Z)) -> times_out_gga(s(X), Y, Z) U1_ga(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_ga(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U2_ga(X, Y, Xs, T, factor_out_ga(cons(Z, Xs), T)) -> factor_out_ga(cons(X, cons(Y, Xs)), T) The argument filtering Pi contains the following mapping: factor_in_ga(x1, x2) = factor_in_ga(x1) cons(x1, x2) = cons(x1, x2) nil = nil factor_out_ga(x1, x2) = factor_out_ga(x2) U1_ga(x1, x2, x3, x4, x5) = U1_ga(x3, x5) times_in_gga(x1, x2, x3) = times_in_gga(x1, x2) 0 = 0 times_out_gga(x1, x2, x3) = times_out_gga(x3) s(x1) = s(x1) U3_gga(x1, x2, x3, x4) = U3_gga(x2, x4) U4_gga(x1, x2, x3, x4) = U4_gga(x4) plus_in_gga(x1, x2, x3) = plus_in_gga(x1, x2) plus_out_gga(x1, x2, x3) = plus_out_gga(x3) U5_gga(x1, x2, x3, x4) = U5_gga(x4) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) FACTOR_IN_GA(x1, x2) = FACTOR_IN_GA(x1) U1_GA(x1, x2, x3, x4, x5) = U1_GA(x3, x5) TIMES_IN_GGA(x1, x2, x3) = TIMES_IN_GGA'(x1, x2) U3_GGA(x1, x2, x3, x4) = U3_GGA(x2, x4) U4_GGA(x1, x2, x3, x4) = U4_GGA(x4) PLUS_IN_GGA(x1, x2, x3) = PLUS_IN_GGA(x1, x2) U5_GGA(x1, x2, x3, x4) = U5_GGA(x4) U2_GA(x1, x2, x3, x4, x5) = U2_GA(x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 6 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Pi DP problem: The TRS P consists of the following rules: PLUS_IN_GGA(s(X), Y, s(Z)) -> PLUS_IN_GGA(X, Y, Z) The TRS R consists of the following rules: factor_in_ga(cons(X, nil), X) -> factor_out_ga(cons(X, nil), X) factor_in_ga(cons(X, cons(Y, Xs)), T) -> U1_ga(X, Y, Xs, T, times_in_gga(X, Y, Z)) times_in_gga(0, X_, 0) -> times_out_gga(0, X_, 0) times_in_gga(s(X), Y, Z) -> U3_gga(X, Y, Z, times_in_gga(X, Y, XY)) U3_gga(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_gga(X, Y, Z, plus_in_gga(XY, Y, Z)) plus_in_gga(0, X, X) -> plus_out_gga(0, X, X) plus_in_gga(s(X), Y, s(Z)) -> U5_gga(X, Y, Z, plus_in_gga(X, Y, Z)) U5_gga(X, Y, Z, plus_out_gga(X, Y, Z)) -> plus_out_gga(s(X), Y, s(Z)) U4_gga(X, Y, Z, plus_out_gga(XY, Y, Z)) -> times_out_gga(s(X), Y, Z) U1_ga(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_ga(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U2_ga(X, Y, Xs, T, factor_out_ga(cons(Z, Xs), T)) -> factor_out_ga(cons(X, cons(Y, Xs)), T) The argument filtering Pi contains the following mapping: factor_in_ga(x1, x2) = factor_in_ga(x1) cons(x1, x2) = cons(x1, x2) nil = nil factor_out_ga(x1, x2) = factor_out_ga(x2) U1_ga(x1, x2, x3, x4, x5) = U1_ga(x3, x5) times_in_gga(x1, x2, x3) = times_in_gga(x1, x2) 0 = 0 times_out_gga(x1, x2, x3) = times_out_gga(x3) s(x1) = s(x1) U3_gga(x1, x2, x3, x4) = U3_gga(x2, x4) U4_gga(x1, x2, x3, x4) = U4_gga(x4) plus_in_gga(x1, x2, x3) = plus_in_gga(x1, x2) plus_out_gga(x1, x2, x3) = plus_out_gga(x3) U5_gga(x1, x2, x3, x4) = U5_gga(x4) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) PLUS_IN_GGA(x1, x2, x3) = PLUS_IN_GGA(x1, x2) 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: PLUS_IN_GGA(s(X), Y, s(Z)) -> PLUS_IN_GGA(X, Y, Z) R is empty. The argument filtering Pi contains the following mapping: s(x1) = s(x1) PLUS_IN_GGA(x1, x2, x3) = PLUS_IN_GGA(x1, x2) 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: PLUS_IN_GGA(s(X), Y) -> PLUS_IN_GGA(X, Y) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (12) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *PLUS_IN_GGA(s(X), Y) -> PLUS_IN_GGA(X, Y) The graph contains the following edges 1 > 1, 2 >= 2 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Pi DP problem: The TRS P consists of the following rules: TIMES_IN_GGA(s(X), Y, Z) -> TIMES_IN_GGA(X, Y, XY) The TRS R consists of the following rules: factor_in_ga(cons(X, nil), X) -> factor_out_ga(cons(X, nil), X) factor_in_ga(cons(X, cons(Y, Xs)), T) -> U1_ga(X, Y, Xs, T, times_in_gga(X, Y, Z)) times_in_gga(0, X_, 0) -> times_out_gga(0, X_, 0) times_in_gga(s(X), Y, Z) -> U3_gga(X, Y, Z, times_in_gga(X, Y, XY)) U3_gga(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_gga(X, Y, Z, plus_in_gga(XY, Y, Z)) plus_in_gga(0, X, X) -> plus_out_gga(0, X, X) plus_in_gga(s(X), Y, s(Z)) -> U5_gga(X, Y, Z, plus_in_gga(X, Y, Z)) U5_gga(X, Y, Z, plus_out_gga(X, Y, Z)) -> plus_out_gga(s(X), Y, s(Z)) U4_gga(X, Y, Z, plus_out_gga(XY, Y, Z)) -> times_out_gga(s(X), Y, Z) U1_ga(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_ga(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U2_ga(X, Y, Xs, T, factor_out_ga(cons(Z, Xs), T)) -> factor_out_ga(cons(X, cons(Y, Xs)), T) The argument filtering Pi contains the following mapping: factor_in_ga(x1, x2) = factor_in_ga(x1) cons(x1, x2) = cons(x1, x2) nil = nil factor_out_ga(x1, x2) = factor_out_ga(x2) U1_ga(x1, x2, x3, x4, x5) = U1_ga(x3, x5) times_in_gga(x1, x2, x3) = times_in_gga(x1, x2) 0 = 0 times_out_gga(x1, x2, x3) = times_out_gga(x3) s(x1) = s(x1) U3_gga(x1, x2, x3, x4) = U3_gga(x2, x4) U4_gga(x1, x2, x3, x4) = U4_gga(x4) plus_in_gga(x1, x2, x3) = plus_in_gga(x1, x2) plus_out_gga(x1, x2, x3) = plus_out_gga(x3) U5_gga(x1, x2, x3, x4) = U5_gga(x4) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) TIMES_IN_GGA(x1, x2, x3) = TIMES_IN_GGA(x1, x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (16) Obligation: Pi DP problem: The TRS P consists of the following rules: TIMES_IN_GGA(s(X), Y, Z) -> TIMES_IN_GGA(X, Y, XY) R is empty. The argument filtering Pi contains the following mapping: s(x1) = s(x1) TIMES_IN_GGA(x1, x2, x3) = TIMES_IN_GGA(x1, x2) We have to consider all (P,R,Pi)-chains ---------------------------------------- (17) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: TIMES_IN_GGA(s(X), Y) -> TIMES_IN_GGA(X, Y) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (19) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *TIMES_IN_GGA(s(X), Y) -> TIMES_IN_GGA(X, Y) The graph contains the following edges 1 > 1, 2 >= 2 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Pi DP problem: The TRS P consists of the following rules: U1_GA(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> FACTOR_IN_GA(cons(Z, Xs), T) FACTOR_IN_GA(cons(X, cons(Y, Xs)), T) -> U1_GA(X, Y, Xs, T, times_in_gga(X, Y, Z)) The TRS R consists of the following rules: factor_in_ga(cons(X, nil), X) -> factor_out_ga(cons(X, nil), X) factor_in_ga(cons(X, cons(Y, Xs)), T) -> U1_ga(X, Y, Xs, T, times_in_gga(X, Y, Z)) times_in_gga(0, X_, 0) -> times_out_gga(0, X_, 0) times_in_gga(s(X), Y, Z) -> U3_gga(X, Y, Z, times_in_gga(X, Y, XY)) U3_gga(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_gga(X, Y, Z, plus_in_gga(XY, Y, Z)) plus_in_gga(0, X, X) -> plus_out_gga(0, X, X) plus_in_gga(s(X), Y, s(Z)) -> U5_gga(X, Y, Z, plus_in_gga(X, Y, Z)) U5_gga(X, Y, Z, plus_out_gga(X, Y, Z)) -> plus_out_gga(s(X), Y, s(Z)) U4_gga(X, Y, Z, plus_out_gga(XY, Y, Z)) -> times_out_gga(s(X), Y, Z) U1_ga(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> U2_ga(X, Y, Xs, T, factor_in_ga(cons(Z, Xs), T)) U2_ga(X, Y, Xs, T, factor_out_ga(cons(Z, Xs), T)) -> factor_out_ga(cons(X, cons(Y, Xs)), T) The argument filtering Pi contains the following mapping: factor_in_ga(x1, x2) = factor_in_ga(x1) cons(x1, x2) = cons(x1, x2) nil = nil factor_out_ga(x1, x2) = factor_out_ga(x2) U1_ga(x1, x2, x3, x4, x5) = U1_ga(x3, x5) times_in_gga(x1, x2, x3) = times_in_gga(x1, x2) 0 = 0 times_out_gga(x1, x2, x3) = times_out_gga(x3) s(x1) = s(x1) U3_gga(x1, x2, x3, x4) = U3_gga(x2, x4) U4_gga(x1, x2, x3, x4) = U4_gga(x4) plus_in_gga(x1, x2, x3) = plus_in_gga(x1, x2) plus_out_gga(x1, x2, x3) = plus_out_gga(x3) U5_gga(x1, x2, x3, x4) = U5_gga(x4) U2_ga(x1, x2, x3, x4, x5) = U2_ga(x5) FACTOR_IN_GA(x1, x2) = FACTOR_IN_GA(x1) U1_GA(x1, x2, x3, x4, x5) = U1_GA(x3, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (22) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (23) Obligation: Pi DP problem: The TRS P consists of the following rules: U1_GA(X, Y, Xs, T, times_out_gga(X, Y, Z)) -> FACTOR_IN_GA(cons(Z, Xs), T) FACTOR_IN_GA(cons(X, cons(Y, Xs)), T) -> U1_GA(X, Y, Xs, T, times_in_gga(X, Y, Z)) The TRS R consists of the following rules: times_in_gga(0, X_, 0) -> times_out_gga(0, X_, 0) times_in_gga(s(X), Y, Z) -> U3_gga(X, Y, Z, times_in_gga(X, Y, XY)) U3_gga(X, Y, Z, times_out_gga(X, Y, XY)) -> U4_gga(X, Y, Z, plus_in_gga(XY, Y, Z)) U4_gga(X, Y, Z, plus_out_gga(XY, Y, Z)) -> times_out_gga(s(X), Y, Z) plus_in_gga(0, X, X) -> plus_out_gga(0, X, X) plus_in_gga(s(X), Y, s(Z)) -> U5_gga(X, Y, Z, plus_in_gga(X, Y, Z)) U5_gga(X, Y, Z, plus_out_gga(X, Y, Z)) -> plus_out_gga(s(X), Y, s(Z)) The argument filtering Pi contains the following mapping: cons(x1, x2) = cons(x1, x2) times_in_gga(x1, x2, x3) = times_in_gga(x1, x2) 0 = 0 times_out_gga(x1, x2, x3) = times_out_gga(x3) s(x1) = s(x1) U3_gga(x1, x2, x3, x4) = U3_gga(x2, x4) U4_gga(x1, x2, x3, x4) = U4_gga(x4) plus_in_gga(x1, x2, x3) = plus_in_gga(x1, x2) plus_out_gga(x1, x2, x3) = plus_out_gga(x3) U5_gga(x1, x2, x3, x4) = U5_gga(x4) FACTOR_IN_GA(x1, x2) = FACTOR_IN_GA(x1) U1_GA(x1, x2, x3, x4, x5) = U1_GA(x3, x5) We have to consider all (P,R,Pi)-chains ---------------------------------------- (24) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: U1_GA(Xs, times_out_gga(Z)) -> FACTOR_IN_GA(cons(Z, Xs)) FACTOR_IN_GA(cons(X, cons(Y, Xs))) -> U1_GA(Xs, times_in_gga(X, Y)) The TRS R consists of the following rules: times_in_gga(0, X_) -> times_out_gga(0) times_in_gga(s(X), Y) -> U3_gga(Y, times_in_gga(X, Y)) U3_gga(Y, times_out_gga(XY)) -> U4_gga(plus_in_gga(XY, Y)) U4_gga(plus_out_gga(Z)) -> times_out_gga(Z) plus_in_gga(0, X) -> plus_out_gga(X) plus_in_gga(s(X), Y) -> U5_gga(plus_in_gga(X, Y)) U5_gga(plus_out_gga(Z)) -> plus_out_gga(s(Z)) The set Q consists of the following terms: times_in_gga(x0, x1) U3_gga(x0, x1) U4_gga(x0) plus_in_gga(x0, x1) U5_gga(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (26) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. U1_GA(Xs, times_out_gga(Z)) -> FACTOR_IN_GA(cons(Z, Xs)) FACTOR_IN_GA(cons(X, cons(Y, Xs))) -> U1_GA(Xs, times_in_gga(X, Y)) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. U1_GA(x1, x2) = U1_GA(x1, x2) times_out_gga(x1) = times_out_gga FACTOR_IN_GA(x1) = FACTOR_IN_GA(x1) cons(x1, x2) = cons(x2) times_in_gga(x1, x2) = times_in_gga 0 = 0 s(x1) = s(x1) U3_gga(x1, x2) = x2 U4_gga(x1) = U4_gga plus_in_gga(x1, x2) = x1 plus_out_gga(x1) = x1 U5_gga(x1) = U5_gga(x1) Recursive path order with status [RPO]. Quasi-Precedence: [U1_GA_2, cons_1] > FACTOR_IN_GA_1 > [times_out_gga, times_in_gga, 0, s_1, U4_gga, U5_gga_1] Status: U1_GA_2: multiset status times_out_gga: [] FACTOR_IN_GA_1: multiset status cons_1: multiset status times_in_gga: [] 0: multiset status s_1: multiset status U4_gga: [] U5_gga_1: [1] The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: times_in_gga(0, X_) -> times_out_gga(0) times_in_gga(s(X), Y) -> U3_gga(Y, times_in_gga(X, Y)) U3_gga(Y, times_out_gga(XY)) -> U4_gga(plus_in_gga(XY, Y)) U4_gga(plus_out_gga(Z)) -> times_out_gga(Z) ---------------------------------------- (27) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: times_in_gga(0, X_) -> times_out_gga(0) times_in_gga(s(X), Y) -> U3_gga(Y, times_in_gga(X, Y)) U3_gga(Y, times_out_gga(XY)) -> U4_gga(plus_in_gga(XY, Y)) U4_gga(plus_out_gga(Z)) -> times_out_gga(Z) plus_in_gga(0, X) -> plus_out_gga(X) plus_in_gga(s(X), Y) -> U5_gga(plus_in_gga(X, Y)) U5_gga(plus_out_gga(Z)) -> plus_out_gga(s(Z)) The set Q consists of the following terms: times_in_gga(x0, x1) U3_gga(x0, x1) U4_gga(x0) plus_in_gga(x0, x1) U5_gga(x0) We have to consider all (P,Q,R)-chains. ---------------------------------------- (28) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (29) YES