/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 17 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, 0 ms] (27) QDP (28) UsableRulesProof [EQUIVALENT, 0 ms] (29) QDP (30) QReductionProof [EQUIVALENT, 0 ms] (31) QDP (32) TransformationProof [EQUIVALENT, 0 ms] (33) QDP (34) TransformationProof [EQUIVALENT, 0 ms] (35) QDP (36) TransformationProof [EQUIVALENT, 0 ms] (37) QDP (38) TransformationProof [EQUIVALENT, 0 ms] (39) QDP (40) TransformationProof [EQUIVALENT, 0 ms] (41) QDP (42) TransformationProof [EQUIVALENT, 0 ms] (43) QDP (44) TransformationProof [EQUIVALENT, 0 ms] (45) QDP (46) TransformationProof [EQUIVALENT, 0 ms] (47) QDP (48) UsableRulesProof [EQUIVALENT, 0 ms] (49) QDP (50) TransformationProof [EQUIVALENT, 0 ms] (51) QDP (52) UsableRulesProof [EQUIVALENT, 0 ms] (53) QDP (54) QReductionProof [EQUIVALENT, 0 ms] (55) QDP (56) RemovalProof [SOUND, 0 ms] (57) QDP (58) NonInfProof [EQUIVALENT, 0 ms] (59) QDP (60) DependencyGraphProof [EQUIVALENT, 0 ms] (61) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1024 -> 1024_1(0) 1024_1(x) -> if(lt(x, 10), x) if(true, x) -> double(1024_1(s(x))) if(false, x) -> s(0) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 double(s(x)) -> s(s(double(x))) 10 -> double(s(double(s(s(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: 1024 -> 1024_1(0) 1024_1(x) -> if(lt(x, 10), x) if(true, x) -> double(1024_1(s(x))) if(false, x) -> s(0) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 double(s(x)) -> s(s(double(x))) 10 -> double(s(double(s(s(0))))) The set Q consists of the following terms: 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 ---------------------------------------- (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: 1024^1 -> 1024_1^1(0) 1024_1^1(x) -> IF(lt(x, 10), x) 1024_1^1(x) -> LT(x, 10) 1024_1^1(x) -> 10^1 IF(true, x) -> DOUBLE(1024_1(s(x))) IF(true, x) -> 1024_1^1(s(x)) LT(s(x), s(y)) -> LT(x, y) DOUBLE(s(x)) -> DOUBLE(x) 10^1 -> DOUBLE(s(double(s(s(0))))) 10^1 -> DOUBLE(s(s(0))) The TRS R consists of the following rules: 1024 -> 1024_1(0) 1024_1(x) -> if(lt(x, 10), x) if(true, x) -> double(1024_1(s(x))) if(false, x) -> s(0) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 double(s(x)) -> s(s(double(x))) 10 -> double(s(double(s(s(0))))) The set Q consists of the following terms: 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 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 6 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: DOUBLE(s(x)) -> DOUBLE(x) The TRS R consists of the following rules: 1024 -> 1024_1(0) 1024_1(x) -> if(lt(x, 10), x) if(true, x) -> double(1024_1(s(x))) if(false, x) -> s(0) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 double(s(x)) -> s(s(double(x))) 10 -> double(s(double(s(s(0))))) The set Q consists of the following terms: 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 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: DOUBLE(s(x)) -> DOUBLE(x) R is empty. The set Q consists of the following terms: 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 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]. 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: DOUBLE(s(x)) -> DOUBLE(x) 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: *DOUBLE(s(x)) -> DOUBLE(x) The graph contains the following edges 1 > 1 ---------------------------------------- (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: 1024 -> 1024_1(0) 1024_1(x) -> if(lt(x, 10), x) if(true, x) -> double(1024_1(s(x))) if(false, x) -> s(0) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 double(s(x)) -> s(s(double(x))) 10 -> double(s(double(s(s(0))))) The set Q consists of the following terms: 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 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: 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 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]. 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 ---------------------------------------- (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) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, 10), x) The TRS R consists of the following rules: 1024 -> 1024_1(0) 1024_1(x) -> if(lt(x, 10), x) if(true, x) -> double(1024_1(s(x))) if(false, x) -> s(0) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 double(s(x)) -> s(s(double(x))) 10 -> double(s(double(s(s(0))))) The set Q consists of the following terms: 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 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) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, 10), x) The TRS R consists of the following rules: 10 -> double(s(double(s(s(0))))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(s(x)) -> s(s(double(x))) double(0) -> 0 The set Q consists of the following terms: 1024 1024_1(x0) if(true, x0) if(false, x0) lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 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]. 1024 1024_1(x0) if(true, x0) if(false, x0) ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, 10), x) The TRS R consists of the following rules: 10 -> double(s(double(s(s(0))))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(s(x)) -> s(s(double(x))) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, 10), x) at position [0,1] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, double(s(double(s(s(0)))))), x),1024_1^1(x) -> IF(lt(x, double(s(double(s(s(0)))))), x)) ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, double(s(double(s(s(0)))))), x) The TRS R consists of the following rules: 10 -> double(s(double(s(s(0))))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(s(x)) -> s(s(double(x))) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) 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. ---------------------------------------- (29) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, double(s(double(s(s(0)))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) 10 We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (30) 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]. 10 ---------------------------------------- (31) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, double(s(double(s(s(0)))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (32) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, double(s(double(s(s(0)))))), x) at position [0,1] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(double(double(s(s(0))))))), x),1024_1^1(x) -> IF(lt(x, s(s(double(double(s(s(0))))))), x)) ---------------------------------------- (33) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(double(double(s(s(0))))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (34) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, s(s(double(double(s(s(0))))))), x) at position [0,1,0,0,0] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(double(s(s(double(s(0)))))))), x),1024_1^1(x) -> IF(lt(x, s(s(double(s(s(double(s(0)))))))), x)) ---------------------------------------- (35) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(double(s(s(double(s(0)))))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, s(s(double(s(s(double(s(0)))))))), x) at position [0,1,0,0] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(s(s(double(s(double(s(0))))))))), x),1024_1^1(x) -> IF(lt(x, s(s(s(s(double(s(double(s(0))))))))), x)) ---------------------------------------- (37) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(double(s(double(s(0))))))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (38) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, s(s(s(s(double(s(double(s(0))))))))), x) at position [0,1,0,0,0,0] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(double(double(s(0)))))))))), x),1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(double(double(s(0)))))))))), x)) ---------------------------------------- (39) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(double(double(s(0)))))))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (40) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(double(double(s(0)))))))))), x) at position [0,1,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(double(s(s(double(0))))))))))), x),1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(double(s(s(double(0))))))))))), x)) ---------------------------------------- (41) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(double(s(s(double(0))))))))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (42) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(double(s(s(double(0))))))))))), x) at position [0,1,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(double(s(double(0)))))))))))), x),1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(double(s(double(0)))))))))))), x)) ---------------------------------------- (43) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(double(s(double(0)))))))))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (44) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(double(s(double(0)))))))))))), x) at position [0,1,0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(double(0))))))))))))), x),1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(double(0))))))))))))), x)) ---------------------------------------- (45) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(double(0))))))))))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (46) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(double(0))))))))))))), x) at position [0,1,0,0,0,0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x),1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x)) ---------------------------------------- (47) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x) The TRS R consists of the following rules: double(s(x)) -> s(s(double(x))) lt(0, s(y)) -> true lt(x, 0) -> false lt(s(x), s(y)) -> lt(x, y) double(0) -> 0 The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (48) 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. ---------------------------------------- (49) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x) The TRS R consists of the following rules: double(0) -> 0 lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) lt(x, 0) -> false The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (50) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(double(0)))))))))))), x) at position [0,1,0,0,0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]: (1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x),1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x)) ---------------------------------------- (51) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x) The TRS R consists of the following rules: double(0) -> 0 lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) lt(x, 0) -> false The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (52) 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. ---------------------------------------- (53) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x) The TRS R consists of the following rules: lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) lt(x, 0) -> false The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) double(0) double(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (54) 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]. double(0) double(s(x0)) ---------------------------------------- (55) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x) -> 1024_1^1(s(x)) 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x) The TRS R consists of the following rules: lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) lt(x, 0) -> false The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (56) RemovalProof (SOUND) In the following pairs the term without variables s(s(s(s(s(s(s(s(s(s(0)))))))))) is replaced by the fresh variable x_removed. Pair: 1024_1^1(x) -> IF(lt(x, s(s(s(s(s(s(s(s(s(s(0))))))))))), x) Positions in right side of the pair: *[0,1]The new variable was added to all pairs as a new argument[CONREM]. ---------------------------------------- (57) Obligation: Q DP problem: The TRS P consists of the following rules: IF(true, x, x_removed) -> 1024_1^1(s(x), x_removed) 1024_1^1(x, x_removed) -> IF(lt(x, x_removed), x, x_removed) The TRS R consists of the following rules: lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) lt(x, 0) -> false The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (58) 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, x_removed) -> 1024_1^1(s(x), x_removed) the following chains were created: *We consider the chain 1024_1^1(x2, x3) -> IF(lt(x2, x3), x2, x3), IF(true, x4, x5) -> 1024_1^1(s(x4), x5) which results in the following constraint: (1) (IF(lt(x2, x3), x2, x3)=IF(true, x4, x5) ==> IF(true, x4, x5)_>=_1024_1^1(s(x4), x5)) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (lt(x2, x3)=true ==> IF(true, x2, x3)_>=_1024_1^1(s(x2), x3)) We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x2, x3)=true which results in the following new constraints: (3) (true=true ==> IF(true, 0, s(x12))_>=_1024_1^1(s(0), s(x12))) (4) (lt(x14, x13)=true & (lt(x14, x13)=true ==> IF(true, x14, x13)_>=_1024_1^1(s(x14), x13)) ==> IF(true, s(x14), s(x13))_>=_1024_1^1(s(s(x14)), s(x13))) We simplified constraint (3) using rules (I), (II) which results in the following new constraint: (5) (IF(true, 0, s(x12))_>=_1024_1^1(s(0), s(x12))) We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (lt(x14, x13)=true ==> IF(true, x14, x13)_>=_1024_1^1(s(x14), x13)) with sigma = [ ] which results in the following new constraint: (6) (IF(true, x14, x13)_>=_1024_1^1(s(x14), x13) ==> IF(true, s(x14), s(x13))_>=_1024_1^1(s(s(x14)), s(x13))) For Pair 1024_1^1(x, x_removed) -> IF(lt(x, x_removed), x, x_removed) the following chains were created: *We consider the chain IF(true, x6, x7) -> 1024_1^1(s(x6), x7), 1024_1^1(x8, x9) -> IF(lt(x8, x9), x8, x9) which results in the following constraint: (1) (1024_1^1(s(x6), x7)=1024_1^1(x8, x9) ==> 1024_1^1(x8, x9)_>=_IF(lt(x8, x9), x8, x9)) We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint: (2) (1024_1^1(s(x6), x7)_>=_IF(lt(s(x6), x7), s(x6), x7)) To summarize, we get the following constraints P__>=_ for the following pairs. *IF(true, x, x_removed) -> 1024_1^1(s(x), x_removed) *(IF(true, 0, s(x12))_>=_1024_1^1(s(0), s(x12))) *(IF(true, x14, x13)_>=_1024_1^1(s(x14), x13) ==> IF(true, s(x14), s(x13))_>=_1024_1^1(s(s(x14)), s(x13))) *1024_1^1(x, x_removed) -> IF(lt(x, x_removed), x, x_removed) *(1024_1^1(s(x6), x7)_>=_IF(lt(s(x6), x7), s(x6), x7)) 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(1024_1^1(x_1, x_2)) = -1 - x_1 + x_2 POL(IF(x_1, x_2, x_3)) = -1 - x_1 - x_2 + x_3 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, x_removed) -> 1024_1^1(s(x), x_removed) The following pairs are in P_bound: IF(true, x, x_removed) -> 1024_1^1(s(x), x_removed) The following rules are usable: true -> lt(0, s(y)) lt(x, y) -> lt(s(x), s(y)) false -> lt(x, 0) ---------------------------------------- (59) Obligation: Q DP problem: The TRS P consists of the following rules: 1024_1^1(x, x_removed) -> IF(lt(x, x_removed), x, x_removed) The TRS R consists of the following rules: lt(0, s(y)) -> true lt(s(x), s(y)) -> lt(x, y) lt(x, 0) -> false The set Q consists of the following terms: lt(0, s(x0)) lt(x0, 0) lt(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (60) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (61) TRUE