/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 96 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 2 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 13 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 0 ms] (10) QTRS (11) QTRSRRRProof [EQUIVALENT, 12 ms] (12) QTRS (13) QTRSRRRProof [EQUIVALENT, 0 ms] (14) QTRS (15) QTRSRRRProof [EQUIVALENT, 0 ms] (16) QTRS (17) QTRSRRRProof [EQUIVALENT, 0 ms] (18) QTRS (19) RisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> a__adx(a__zeros) a__zeros -> cons(0, zeros) a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) a__hd(cons(X, Y)) -> mark(X) a__tl(cons(X, Y)) -> mark(Y) mark(nats) -> a__nats mark(adx(X)) -> a__adx(mark(X)) mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(hd(X)) -> a__hd(mark(X)) mark(tl(X)) -> a__tl(mark(X)) mark(cons(X1, X2)) -> cons(X1, X2) mark(0) -> 0 mark(s(X)) -> s(X) a__nats -> nats a__adx(X) -> adx(X) a__zeros -> zeros a__incr(X) -> incr(X) a__hd(X) -> hd(X) a__tl(X) -> tl(X) The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = x_1 POL(a__hd(x_1)) = 1 + 2*x_1 POL(a__incr(x_1)) = x_1 POL(a__nats) = 0 POL(a__tl(x_1)) = 1 + 2*x_1 POL(a__zeros) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(hd(x_1)) = 1 + 2*x_1 POL(incr(x_1)) = x_1 POL(mark(x_1)) = 2*x_1 POL(nats) = 0 POL(s(x_1)) = x_1 POL(tl(x_1)) = 1 + 2*x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__hd(cons(X, Y)) -> mark(X) a__tl(cons(X, Y)) -> mark(Y) mark(hd(X)) -> a__hd(mark(X)) mark(tl(X)) -> a__tl(mark(X)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> a__adx(a__zeros) a__zeros -> cons(0, zeros) a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) mark(nats) -> a__nats mark(adx(X)) -> a__adx(mark(X)) mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(cons(X1, X2)) -> cons(X1, X2) mark(0) -> 0 mark(s(X)) -> s(X) a__nats -> nats a__adx(X) -> adx(X) a__zeros -> zeros a__incr(X) -> incr(X) a__hd(X) -> hd(X) a__tl(X) -> tl(X) The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = x_1 POL(a__hd(x_1)) = 2 + 2*x_1 POL(a__incr(x_1)) = x_1 POL(a__nats) = 0 POL(a__tl(x_1)) = 1 + 2*x_1 POL(a__zeros) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(hd(x_1)) = 2 + 2*x_1 POL(incr(x_1)) = x_1 POL(mark(x_1)) = 2*x_1 POL(nats) = 0 POL(s(x_1)) = x_1 POL(tl(x_1)) = 2*x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__tl(X) -> tl(X) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> a__adx(a__zeros) a__zeros -> cons(0, zeros) a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) mark(nats) -> a__nats mark(adx(X)) -> a__adx(mark(X)) mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(cons(X1, X2)) -> cons(X1, X2) mark(0) -> 0 mark(s(X)) -> s(X) a__nats -> nats a__adx(X) -> adx(X) a__zeros -> zeros a__incr(X) -> incr(X) a__hd(X) -> hd(X) The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = x_1 POL(a__hd(x_1)) = 1 + x_1 POL(a__incr(x_1)) = x_1 POL(a__nats) = 0 POL(a__zeros) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(hd(x_1)) = x_1 POL(incr(x_1)) = x_1 POL(mark(x_1)) = 2*x_1 POL(nats) = 0 POL(s(x_1)) = x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__hd(X) -> hd(X) ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__nats -> a__adx(a__zeros) a__zeros -> cons(0, zeros) a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) mark(nats) -> a__nats mark(adx(X)) -> a__adx(mark(X)) mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(cons(X1, X2)) -> cons(X1, X2) mark(0) -> 0 mark(s(X)) -> s(X) a__nats -> nats a__adx(X) -> adx(X) a__zeros -> zeros a__incr(X) -> incr(X) The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (7) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = x_1 POL(a__incr(x_1)) = x_1 POL(a__nats) = 1 POL(a__zeros) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(incr(x_1)) = x_1 POL(mark(x_1)) = 2*x_1 POL(nats) = 1 POL(s(x_1)) = x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__nats -> a__adx(a__zeros) mark(nats) -> a__nats ---------------------------------------- (8) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) mark(adx(X)) -> a__adx(mark(X)) mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(cons(X1, X2)) -> cons(X1, X2) mark(0) -> 0 mark(s(X)) -> s(X) a__nats -> nats a__adx(X) -> adx(X) a__zeros -> zeros a__incr(X) -> incr(X) The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (9) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = x_1 POL(a__incr(x_1)) = x_1 POL(a__nats) = 1 POL(a__zeros) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(incr(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(nats) = 0 POL(s(x_1)) = x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__nats -> nats ---------------------------------------- (10) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) mark(adx(X)) -> a__adx(mark(X)) mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(cons(X1, X2)) -> cons(X1, X2) mark(0) -> 0 mark(s(X)) -> s(X) a__adx(X) -> adx(X) a__zeros -> zeros a__incr(X) -> incr(X) The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (11) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = 2 + 2*x_1 POL(a__incr(x_1)) = 2 + x_1 POL(a__zeros) = 2 POL(adx(x_1)) = 1 + 2*x_1 POL(cons(x_1, x_2)) = 1 + 2*x_1 + x_2 POL(incr(x_1)) = 1 + x_1 POL(mark(x_1)) = 2*x_1 POL(s(x_1)) = x_1 POL(zeros) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__incr(cons(X, Y)) -> cons(s(X), incr(Y)) mark(cons(X1, X2)) -> cons(X1, X2) a__adx(X) -> adx(X) a__zeros -> zeros a__incr(X) -> incr(X) ---------------------------------------- (12) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) mark(adx(X)) -> a__adx(mark(X)) mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(0) -> 0 mark(s(X)) -> s(X) The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (13) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = 2 + x_1 POL(a__incr(x_1)) = 1 + x_1 POL(a__zeros) = 1 POL(adx(x_1)) = 1 + x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(incr(x_1)) = 2 + 2*x_1 POL(mark(x_1)) = 2*x_1 POL(s(x_1)) = 1 + 2*x_1 POL(zeros) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(zeros) -> a__zeros mark(incr(X)) -> a__incr(mark(X)) mark(s(X)) -> s(X) ---------------------------------------- (14) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) mark(adx(X)) -> a__adx(mark(X)) mark(0) -> 0 The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (15) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__adx(x_1)) = 2*x_1 POL(a__incr(x_1)) = x_1 POL(a__zeros) = 2 POL(adx(x_1)) = 2 + 2*x_1 POL(cons(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(mark(x_1)) = 1 + 2*x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(adx(X)) -> a__adx(mark(X)) mark(0) -> 0 ---------------------------------------- (16) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (17) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:adx_1 > a__adx_1 > a__incr_1 > zeros > 0 > a__zeros > cons_2 and weight map: a__zeros=2 0=1 zeros=1 a__adx_1=2 a__incr_1=1 adx_1=1 cons_2=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__zeros -> cons(0, zeros) a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y))) ---------------------------------------- (18) Obligation: Q restricted rewrite system: R is empty. The set Q consists of the following terms: a__nats a__zeros mark(nats) mark(adx(x0)) mark(zeros) mark(incr(x0)) mark(hd(x0)) mark(tl(x0)) mark(cons(x0, x1)) mark(0) mark(s(x0)) a__adx(x0) a__incr(x0) a__hd(x0) a__tl(x0) ---------------------------------------- (19) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (20) YES