/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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) NonInfProof [EQUIVALENT, 46 ms] (27) QDP (28) DependencyGraphProof [EQUIVALENT, 0 ms] (29) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: times(x, y) -> help(x, y, 0) help(x, y, c) -> if(lt(c, y), x, y, c) if(true, x, y, c) -> plus(x, help(x, y, s(c))) if(false, x, y, c) -> 0 lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(0, x) -> x plus(x, s(y)) -> s(plus(x, y)) plus(s(x), y) -> s(plus(x, y)) 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: times(x, y) -> help(x, y, 0) help(x, y, c) -> if(lt(c, y), x, y, c) if(true, x, y, c) -> plus(x, help(x, y, s(c))) if(false, x, y, c) -> 0 lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(0, x) -> x plus(x, s(y)) -> s(plus(x, y)) plus(s(x), y) -> s(plus(x, y)) The set Q consists of the following terms: times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) ---------------------------------------- (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: TIMES(x, y) -> HELP(x, y, 0) HELP(x, y, c) -> IF(lt(c, y), x, y, c) HELP(x, y, c) -> LT(c, y) IF(true, x, y, c) -> PLUS(x, help(x, y, s(c))) IF(true, x, y, c) -> HELP(x, y, s(c)) LT(s(x), s(y)) -> LT(x, y) PLUS(x, s(y)) -> PLUS(x, y) PLUS(s(x), y) -> PLUS(x, y) The TRS R consists of the following rules: times(x, y) -> help(x, y, 0) help(x, y, c) -> if(lt(c, y), x, y, c) if(true, x, y, c) -> plus(x, help(x, y, s(c))) if(false, x, y, c) -> 0 lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(0, x) -> x plus(x, s(y)) -> s(plus(x, y)) plus(s(x), y) -> s(plus(x, y)) The set Q consists of the following terms: times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) 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(s(x), y) -> PLUS(x, y) PLUS(x, s(y)) -> PLUS(x, y) The TRS R consists of the following rules: times(x, y) -> help(x, y, 0) help(x, y, c) -> if(lt(c, y), x, y, c) if(true, x, y, c) -> plus(x, help(x, y, s(c))) if(false, x, y, c) -> 0 lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(0, x) -> x plus(x, s(y)) -> s(plus(x, y)) plus(s(x), y) -> s(plus(x, y)) The set Q consists of the following terms: times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) 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(s(x), y) -> PLUS(x, y) PLUS(x, s(y)) -> PLUS(x, y) R is empty. The set Q consists of the following terms: times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) 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]. times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: PLUS(s(x), y) -> PLUS(x, y) 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(s(x), y) -> PLUS(x, y) The graph contains the following edges 1 > 1, 2 >= 2 *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: times(x, y) -> help(x, y, 0) help(x, y, c) -> if(lt(c, y), x, y, c) if(true, x, y, c) -> plus(x, help(x, y, s(c))) if(false, x, y, c) -> 0 lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(0, x) -> x plus(x, s(y)) -> s(plus(x, y)) plus(s(x), y) -> s(plus(x, y)) The set Q consists of the following terms: times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) 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: times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) 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]. times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) ---------------------------------------- (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: IF(true, x, y, c) -> HELP(x, y, s(c)) HELP(x, y, c) -> IF(lt(c, y), x, y, c) The TRS R consists of the following rules: times(x, y) -> help(x, y, 0) help(x, y, c) -> if(lt(c, y), x, y, c) if(true, x, y, c) -> plus(x, help(x, y, s(c))) if(false, x, y, c) -> 0 lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) plus(x, 0) -> x plus(0, x) -> x plus(x, s(y)) -> s(plus(x, y)) plus(s(x), y) -> s(plus(x, y)) The set Q consists of the following terms: times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) 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: IF(true, x, y, c) -> HELP(x, y, s(c)) HELP(x, y, c) -> IF(lt(c, y), x, y, c) The TRS R consists of the following rules: lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) The set Q consists of the following terms: times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) 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]. times(x0, x1) help(x0, x1, x2) if(true, x0, x1, x2) if(false, x0, x1, x2) plus(x0, 0) plus(0, x0) plus(x0, s(x1)) plus(s(x0), x1) ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x, y, c) -> HELP(x, y, s(c)) HELP(x, y, c) -> IF(lt(c, y), x, y, c) The TRS R consists of the following rules: lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) The set Q consists of the following terms: lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) 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 IF(true, x, y, c) -> HELP(x, y, s(c)) the following chains were created: *We consider the chain HELP(x3, x4, x5) -> IF(lt(x5, x4), x3, x4, x5), IF(true, x6, x7, x8) -> HELP(x6, x7, s(x8)) which results in the following constraint: (1) (IF(lt(x5, x4), x3, x4, x5)=IF(true, x6, x7, x8) ==> IF(true, x6, x7, x8)_>=_HELP(x6, x7, s(x8))) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (lt(x5, x4)=true ==> IF(true, x3, x4, x5)_>=_HELP(x3, x4, s(x5))) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x5, x4)=true which results in the following new constraints: (3) (true=true ==> IF(true, x3, s(x18), 0)_>=_HELP(x3, s(x18), s(0))) (4) (lt(x21, x20)=true & (\/x22:lt(x21, x20)=true ==> IF(true, x22, x20, x21)_>=_HELP(x22, x20, s(x21))) ==> IF(true, x3, s(x20), s(x21))_>=_HELP(x3, s(x20), s(s(x21)))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (IF(true, x3, s(x18), 0)_>=_HELP(x3, s(x18), s(0))) We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (\/x22:lt(x21, x20)=true ==> IF(true, x22, x20, x21)_>=_HELP(x22, x20, s(x21))) with sigma = [x22 / x3] which results in the following new constraint: (6) (IF(true, x3, x20, x21)_>=_HELP(x3, x20, s(x21)) ==> IF(true, x3, s(x20), s(x21))_>=_HELP(x3, s(x20), s(s(x21)))) For Pair HELP(x, y, c) -> IF(lt(c, y), x, y, c) the following chains were created: *We consider the chain IF(true, x9, x10, x11) -> HELP(x9, x10, s(x11)), HELP(x12, x13, x14) -> IF(lt(x14, x13), x12, x13, x14) which results in the following constraint: (1) (HELP(x9, x10, s(x11))=HELP(x12, x13, x14) ==> HELP(x12, x13, x14)_>=_IF(lt(x14, x13), x12, x13, x14)) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (HELP(x9, x10, s(x11))_>=_IF(lt(s(x11), x10), x9, x10, s(x11))) To summarize, we get the following constraints P__>=_ for the following pairs. *IF(true, x, y, c) -> HELP(x, y, s(c)) *(IF(true, x3, s(x18), 0)_>=_HELP(x3, s(x18), s(0))) *(IF(true, x3, x20, x21)_>=_HELP(x3, x20, s(x21)) ==> IF(true, x3, s(x20), s(x21))_>=_HELP(x3, s(x20), s(s(x21)))) *HELP(x, y, c) -> IF(lt(c, y), x, y, c) *(HELP(x9, x10, s(x11))_>=_IF(lt(s(x11), x10), x9, x10, s(x11))) 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) = -2 POL(false) = 0 POL(lt(x_1, x_2)) = 0 POL(s(x_1)) = 1 + x_1 POL(true) = 0 The following pairs are in P_>: IF(true, x, y, c) -> HELP(x, y, s(c)) The following pairs are in P_bound: IF(true, x, y, c) -> HELP(x, y, s(c)) The following rules are usable: true -> lt(0, s(x)) false -> lt(s(x), 0) lt(x, y) -> lt(s(x), s(y)) ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: HELP(x, y, c) -> IF(lt(c, y), x, y, c) The TRS R consists of the following rules: lt(0, s(x)) -> true lt(s(x), 0) -> false lt(s(x), s(y)) -> lt(x, y) The set Q consists of the following terms: lt(0, s(x0)) lt(s(x0), 0) lt(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (29) TRUE