details

property	value
status	complete
benchmark	gen-1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n167.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.65333390236 seconds
cpu usage	7.218167343
max memory	5.01972992E8

stage attributes

key	value
output-size	5900
starexec-result	YES

output

/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 85 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z 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: F(c(a, z, x)) -> B(a, z) B(x, b(z, y)) -> F(b(f(f(z)), c(x, z, y))) B(x, b(z, y)) -> B(f(f(z)), c(x, z, y)) B(x, b(z, y)) -> F(f(z)) B(x, b(z, y)) -> F(z) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z 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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B(x, b(z, y)) -> F(b(f(f(z)), c(x, z, y))) F(c(a, z, x)) -> B(a, z) B(x, b(z, y)) -> F(f(z)) B(x, b(z, y)) -> F(z) The TRS R consists of the following rules: f(c(a, z, x)) -> b(a, z) b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) b(y, z) -> z Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule B(x, b(z, y)) -> F(b(f(f(z)), c(x, z, y))) we obtained the following new rules [LPAR04]: (B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2))),B(a, b(x1, x2)) -> F(b(f(f(x1)), c(a, x1, x2)))) popout

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job log

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actions

all output return to TRS Standard