/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 675 ms] (6) QDP (7) QDPSizeChangeProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, p(1, a(y, t)))) a(p(x, y), z) -> p(a(x, z), a(y, z)) a(a(x, y), z) -> a(x, a(y, z)) lambda(x) -> x a(x, y) -> x a(x, y) -> y p(x, y) -> x p(x, y) -> y Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: A(lambda(x), y) -> LAMBDA(a(x, p(1, a(y, t)))) A(lambda(x), y) -> A(x, p(1, a(y, t))) A(lambda(x), y) -> P(1, a(y, t)) A(lambda(x), y) -> A(y, t) A(p(x, y), z) -> P(a(x, z), a(y, z)) A(p(x, y), z) -> A(x, z) A(p(x, y), z) -> A(y, z) A(a(x, y), z) -> A(x, a(y, z)) A(a(x, y), z) -> A(y, z) The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, p(1, a(y, t)))) a(p(x, y), z) -> p(a(x, z), a(y, z)) a(a(x, y), z) -> a(x, a(y, z)) lambda(x) -> x a(x, y) -> x a(x, y) -> y p(x, y) -> x p(x, y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(lambda(x), y) -> A(y, t) A(lambda(x), y) -> A(x, p(1, a(y, t))) A(p(x, y), z) -> A(x, z) A(p(x, y), z) -> A(y, z) A(a(x, y), z) -> A(x, a(y, z)) A(a(x, y), z) -> A(y, z) The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, p(1, a(y, t)))) a(p(x, y), z) -> p(a(x, z), a(y, z)) a(a(x, y), z) -> a(x, a(y, z)) lambda(x) -> x a(x, y) -> x a(x, y) -> y p(x, y) -> x p(x, y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A(lambda(x), y) -> A(y, t) A(lambda(x), y) -> A(x, p(1, a(y, t))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation with max and min functions [POLO,MAXPOLO]: POL(1) = 0 POL(A(x_1, x_2)) = x_1 + x_2 POL(a(x_1, x_2)) = x_1 + x_2 POL(lambda(x_1)) = 1 + x_1 POL(p(x_1, x_2)) = max(x_1, x_2) POL(t) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: a(lambda(x), y) -> lambda(a(x, p(1, a(y, t)))) a(p(x, y), z) -> p(a(x, z), a(y, z)) a(a(x, y), z) -> a(x, a(y, z)) a(x, y) -> x a(x, y) -> y p(x, y) -> x p(x, y) -> y lambda(x) -> x ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(p(x, y), z) -> A(x, z) A(p(x, y), z) -> A(y, z) A(a(x, y), z) -> A(x, a(y, z)) A(a(x, y), z) -> A(y, z) The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, p(1, a(y, t)))) a(p(x, y), z) -> p(a(x, z), a(y, z)) a(a(x, y), z) -> a(x, a(y, z)) lambda(x) -> x a(x, y) -> x a(x, y) -> y p(x, y) -> x p(x, y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) 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: *A(p(x, y), z) -> A(x, z) The graph contains the following edges 1 > 1, 2 >= 2 *A(p(x, y), z) -> A(y, z) The graph contains the following edges 1 > 1, 2 >= 2 *A(a(x, y), z) -> A(x, a(y, z)) The graph contains the following edges 1 > 1 *A(a(x, y), z) -> A(y, z) The graph contains the following edges 1 > 1, 2 >= 2 ---------------------------------------- (8) YES