/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) QReductionProof [EQUIVALENT, 0 ms] (25) QDP (26) TransformationProof [EQUIVALENT, 3 ms] (27) QDP (28) NonInfProof [EQUIVALENT, 0 ms] (29) QDP (30) DependencyGraphProof [EQUIVALENT, 0 ms] (31) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) quot(x, s(y)) -> help(x, s(y), 0) help(x, s(y), c) -> if(lt(c, x), x, s(y), c) if(true, x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) if(false, x, s(y), c) -> 0 Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) quot(x, s(y)) -> help(x, s(y), 0) help(x, s(y), c) -> if(lt(c, x), x, s(y), c) if(true, x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) if(false, x, s(y), c) -> 0 The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: LT(s(x), s(y)) -> LT(x, y) PLUS(x, s(y)) -> PLUS(x, y) QUOT(x, s(y)) -> HELP(x, s(y), 0) HELP(x, s(y), c) -> IF(lt(c, x), x, s(y), c) HELP(x, s(y), c) -> LT(c, x) IF(true, x, s(y), c) -> HELP(x, s(y), plus(c, s(y))) IF(true, x, s(y), c) -> PLUS(c, s(y)) The TRS R consists of the following rules: lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) quot(x, s(y)) -> help(x, s(y), 0) help(x, s(y), c) -> if(lt(c, x), x, s(y), c) if(true, x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) if(false, x, s(y), c) -> 0 The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(x, s(y)) -> PLUS(x, y) The TRS R consists of the following rules: lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) quot(x, s(y)) -> help(x, s(y), 0) help(x, s(y), c) -> if(lt(c, x), x, s(y), c) if(true, x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) if(false, x, s(y), c) -> 0 The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(x, s(y)) -> PLUS(x, y) R is empty. The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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]. lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(x, s(y)) -> PLUS(x, y) R is empty. Q is empty. We have to consider all minimal (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(x, s(y)) -> PLUS(x, y) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: LT(s(x), s(y)) -> LT(x, y) The TRS R consists of the following rules: lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) quot(x, s(y)) -> help(x, s(y), 0) help(x, s(y), c) -> if(lt(c, x), x, s(y), c) if(true, x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) if(false, x, s(y), c) -> 0 The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: LT(s(x), s(y)) -> LT(x, y) R is empty. The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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]. lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: LT(s(x), s(y)) -> LT(x, y) R is empty. Q is empty. We have to consider all minimal (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: *LT(s(x), s(y)) -> LT(x, y) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (20) YES ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: HELP(x, s(y), c) -> IF(lt(c, x), x, s(y), c) IF(true, x, s(y), c) -> HELP(x, s(y), plus(c, s(y))) The TRS R consists of the following rules: lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) quot(x, s(y)) -> help(x, s(y), 0) help(x, s(y), c) -> if(lt(c, x), x, s(y), c) if(true, x, s(y), c) -> s(help(x, s(y), plus(c, s(y)))) if(false, x, s(y), c) -> 0 The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) 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. ---------------------------------------- (23) Obligation: Q DP problem: The TRS P consists of the following rules: HELP(x, s(y), c) -> IF(lt(c, x), x, s(y), c) IF(true, x, s(y), c) -> HELP(x, s(y), plus(c, s(y))) The TRS R consists of the following rules: plus(x, s(y)) -> s(plus(x, y)) plus(x, 0) -> x lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) 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]. quot(x0, s(x1)) help(x0, s(x1), x2) if(true, x0, s(x1), x2) if(false, x0, s(x1), x2) ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: HELP(x, s(y), c) -> IF(lt(c, x), x, s(y), c) IF(true, x, s(y), c) -> HELP(x, s(y), plus(c, s(y))) The TRS R consists of the following rules: plus(x, s(y)) -> s(plus(x, y)) plus(x, 0) -> x lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule IF(true, x, s(y), c) -> HELP(x, s(y), plus(c, s(y))) at position [2] we obtained the following new rules [LPAR04]: (IF(true, x, s(y), c) -> HELP(x, s(y), s(plus(c, y))),IF(true, x, s(y), c) -> HELP(x, s(y), s(plus(c, y)))) ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: HELP(x, s(y), c) -> IF(lt(c, x), x, s(y), c) IF(true, x, s(y), c) -> HELP(x, s(y), s(plus(c, y))) The TRS R consists of the following rules: plus(x, s(y)) -> s(plus(x, y)) plus(x, 0) -> x lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) NonInfProof (EQUIVALENT) The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: Note that final constraints are written in bold face. For Pair HELP(x, s(y), c) -> IF(lt(c, x), x, s(y), c) the following chains were created: *We consider the chain IF(true, x3, s(x4), x5) -> HELP(x3, s(x4), s(plus(x5, x4))), HELP(x6, s(x7), x8) -> IF(lt(x8, x6), x6, s(x7), x8) which results in the following constraint: (1) (HELP(x3, s(x4), s(plus(x5, x4)))=HELP(x6, s(x7), x8) ==> HELP(x6, s(x7), x8)_>=_IF(lt(x8, x6), x6, s(x7), x8)) We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint: (2) (HELP(x3, s(x4), x8)_>=_IF(lt(x8, x3), x3, s(x4), x8)) For Pair IF(true, x, s(y), c) -> HELP(x, s(y), s(plus(c, y))) the following chains were created: *We consider the chain HELP(x9, s(x10), x11) -> IF(lt(x11, x9), x9, s(x10), x11), IF(true, x12, s(x13), x14) -> HELP(x12, s(x13), s(plus(x14, x13))) which results in the following constraint: (1) (IF(lt(x11, x9), x9, s(x10), x11)=IF(true, x12, s(x13), x14) ==> IF(true, x12, s(x13), x14)_>=_HELP(x12, s(x13), s(plus(x14, x13)))) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (lt(x11, x9)=true ==> IF(true, x9, s(x10), x11)_>=_HELP(x9, s(x10), s(plus(x11, x10)))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x11, x9)=true which results in the following new constraints: (3) (true=true ==> IF(true, s(x19), s(x10), 0)_>=_HELP(s(x19), s(x10), s(plus(0, x10)))) (4) (lt(x21, x20)=true & (\/x22:lt(x21, x20)=true ==> IF(true, x20, s(x22), x21)_>=_HELP(x20, s(x22), s(plus(x21, x22)))) ==> IF(true, s(x20), s(x10), s(x21))_>=_HELP(s(x20), s(x10), s(plus(s(x21), x10)))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (IF(true, s(x19), s(x10), 0)_>=_HELP(s(x19), s(x10), s(plus(0, x10)))) We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (\/x22:lt(x21, x20)=true ==> IF(true, x20, s(x22), x21)_>=_HELP(x20, s(x22), s(plus(x21, x22)))) with sigma = [x22 / x10] which results in the following new constraint: (6) (IF(true, x20, s(x10), x21)_>=_HELP(x20, s(x10), s(plus(x21, x10))) ==> IF(true, s(x20), s(x10), s(x21))_>=_HELP(s(x20), s(x10), s(plus(s(x21), x10)))) To summarize, we get the following constraints P__>=_ for the following pairs. *HELP(x, s(y), c) -> IF(lt(c, x), x, s(y), c) *(HELP(x3, s(x4), x8)_>=_IF(lt(x8, x3), x3, s(x4), x8)) *IF(true, x, s(y), c) -> HELP(x, s(y), s(plus(c, y))) *(IF(true, s(x19), s(x10), 0)_>=_HELP(s(x19), s(x10), s(plus(0, x10)))) *(IF(true, x20, s(x10), x21)_>=_HELP(x20, s(x10), s(plus(x21, x10))) ==> IF(true, s(x20), s(x10), s(x21))_>=_HELP(s(x20), s(x10), s(plus(s(x21), x10)))) The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. Using the following integer polynomial ordering the resulting constraints can be solved Polynomial interpretation [NONINF]: POL(0) = 0 POL(HELP(x_1, x_2, x_3)) = -1 + x_1 + x_2 - x_3 POL(IF(x_1, x_2, x_3, x_4)) = -1 - x_1 + x_2 + x_3 - x_4 POL(c) = -1 POL(false) = 0 POL(lt(x_1, x_2)) = 0 POL(plus(x_1, x_2)) = x_1 POL(s(x_1)) = 1 + x_1 POL(true) = 0 The following pairs are in P_>: IF(true, x, s(y), c) -> HELP(x, s(y), s(plus(c, y))) The following pairs are in P_bound: IF(true, x, s(y), c) -> HELP(x, s(y), s(plus(c, y))) The following rules are usable: false -> lt(x, 0) true -> lt(0, s(y)) lt(x, y) -> lt(s(x), s(y)) s(plus(x, y)) -> plus(x, s(y)) x -> plus(x, 0) ---------------------------------------- (29) Obligation: Q DP problem: The TRS P consists of the following rules: HELP(x, s(y), c) -> IF(lt(c, x), x, s(y), c) The TRS R consists of the following rules: plus(x, s(y)) -> s(plus(x, y)) plus(x, 0) -> x lt(x, 0) -> false lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) The set Q consists of the following terms: lt(x0, 0) lt(0, s(x0)) lt(s(x0), s(x1)) plus(x0, 0) plus(x0, s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (30) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (31) TRUE