/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT,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) MNOCProof [EQUIVALENT, 0 ms] (16) QDP (17) NonLoopProof [COMPLETE, 114 ms] (18) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(tt, x, y) -> g(f(x, y), s(x), s(y)) f(s(x), y) -> f(x, y) f(x, s(y)) -> f(x, y) f(0, 0) -> tt Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is f(s(x), y) -> f(x, y) f(x, s(y)) -> f(x, y) f(0, 0) -> tt The TRS R 2 is g(tt, x, y) -> g(f(x, y), s(x), s(y)) The signature Sigma is {g_3} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(tt, x, y) -> g(f(x, y), s(x), s(y)) f(s(x), y) -> f(x, y) f(x, s(y)) -> f(x, y) f(0, 0) -> tt The set Q consists of the following terms: g(tt, x0, x1) f(s(x0), x1) f(x0, s(x1)) f(0, 0) ---------------------------------------- (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: G(tt, x, y) -> G(f(x, y), s(x), s(y)) G(tt, x, y) -> F(x, y) F(s(x), y) -> F(x, y) F(x, s(y)) -> F(x, y) The TRS R consists of the following rules: g(tt, x, y) -> g(f(x, y), s(x), s(y)) f(s(x), y) -> f(x, y) f(x, s(y)) -> f(x, y) f(0, 0) -> tt The set Q consists of the following terms: g(tt, x0, x1) f(s(x0), x1) f(x0, s(x1)) f(0, 0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, s(y)) -> F(x, y) F(s(x), y) -> F(x, y) The TRS R consists of the following rules: g(tt, x, y) -> g(f(x, y), s(x), s(y)) f(s(x), y) -> f(x, y) f(x, s(y)) -> f(x, y) f(0, 0) -> tt The set Q consists of the following terms: g(tt, x0, x1) f(s(x0), x1) f(x0, s(x1)) f(0, 0) 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: F(x, s(y)) -> F(x, y) F(s(x), y) -> F(x, y) R is empty. The set Q consists of the following terms: g(tt, x0, x1) f(s(x0), x1) f(x0, s(x1)) f(0, 0) 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]. g(tt, x0, x1) f(s(x0), x1) f(x0, s(x1)) f(0, 0) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, s(y)) -> F(x, y) F(s(x), y) -> F(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: *F(x, s(y)) -> F(x, y) The graph contains the following edges 1 >= 1, 2 > 2 *F(s(x), y) -> F(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: G(tt, x, y) -> G(f(x, y), s(x), s(y)) The TRS R consists of the following rules: g(tt, x, y) -> g(f(x, y), s(x), s(y)) f(s(x), y) -> f(x, y) f(x, s(y)) -> f(x, y) f(0, 0) -> tt The set Q consists of the following terms: g(tt, x0, x1) f(s(x0), x1) f(x0, s(x1)) f(0, 0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) MNOCProof (EQUIVALENT) We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: G(tt, x, y) -> G(f(x, y), s(x), s(y)) The TRS R consists of the following rules: g(tt, x, y) -> g(f(x, y), s(x), s(y)) f(s(x), y) -> f(x, y) f(x, s(y)) -> f(x, y) f(0, 0) -> tt Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (17) NonLoopProof (COMPLETE) By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP. We apply the theorem with m = 1, b = 1, σ' = [ ], and μ' = [ ] on the rule G(tt, s(zr0), s(zr2))[zr0 / s(zr0), zr2 / s(zr2)]^n[zr0 / 0, zr2 / 0] -> G(tt, s(s(zr0)), s(s(zr2)))[zr0 / s(zr0), zr2 / s(zr2)]^n[zr0 / 0, zr2 / 0] This rule is correct for the QDP as the following derivation shows: G(tt, s(zr0), s(zr2))[zr0 / s(zr0), zr2 / s(zr2)]^n[zr0 / 0, zr2 / 0] -> G(tt, s(s(zr0)), s(s(zr2)))[zr0 / s(zr0), zr2 / s(zr2)]^n[zr0 / 0, zr2 / 0] by Equivalence by Domain Renaming of the lhs with [zl0 / zr0, zl2 / zr2] intermediate steps: Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) G(tt, s(zl1), s(zl3))[zr1 / s(zr1), zr3 / s(zr3), zl1 / s(zl1), zl3 / s(zl3)]^n[zr1 / 0, zr3 / 0, zl1 / 0, zl3 / 0] -> G(tt, s(s(zr1)), s(s(zr3)))[zr1 / s(zr1), zr3 / s(zr3), zl1 / s(zl1), zl3 / s(zl3)]^n[zr1 / 0, zr3 / 0, zl1 / 0, zl3 / 0] by Narrowing at position: [0] intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) G(tt, s(zl1), s(zs1))[zr1 / s(zr1), zl1 / s(zl1), zs1 / s(zs1)]^n[zr1 / x0, zl1 / x0, zs1 / y0] -> G(f(x0, y0), s(s(zr1)), s(s(zs1)))[zr1 / s(zr1), zl1 / s(zl1), zs1 / s(zs1)]^n[zr1 / x0, zl1 / x0, zs1 / y0] by Narrowing at position: [0] intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate mu - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiate Sigma - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) G(tt, s(zs1), x1)[zs1 / s(zs1)]^n[zs1 / y1] -> G(f(y1, x1), s(s(zs1)), s(x1))[zs1 / s(zs1)]^n[zs1 / y1] by Narrowing at position: [0] intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation G(tt, x, y)[ ]^n[ ] -> G(f(x, y), s(x), s(y))[ ]^n[ ] by Rule from TRS P intermediate steps: Equiv IPS (rhs) - Equiv IPS (rhs) - Instantiation - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (lhs) f(s(x), y)[x / s(x)]^n[ ] -> f(x, y)[ ]^n[ ] by PatternCreation I with delta: [ ], theta: [ ], sigma: [x / s(x)] f(s(x), y)[ ]^n[ ] -> f(x, y)[ ]^n[ ] by Rule from TRS R intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (lhs) f(x, s(y))[y / s(y)]^n[ ] -> f(x, y)[ ]^n[ ] by PatternCreation I with delta: [ ], theta: [ ], sigma: [y / s(y)] f(x, s(y))[ ]^n[ ] -> f(x, y)[ ]^n[ ] by Rule from TRS R intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) f(0, 0)[ ]^n[ ] -> tt[ ]^n[ ] by Rule from TRS R ---------------------------------------- (18) NO