details

property	value
status	complete
benchmark	gen-15.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n053.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.50507688522 seconds
cpu usage	6.673474198
max memory	4.86363136E8

stage attributes

key	value
output-size	6093
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 3 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 31 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(z, x, a) -> f(b(b(f(z), z), x)) b(y, b(z, a)) -> f(b(c(f(a), y, z), z)) f(c(c(z, a, a), x, a)) -> 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: C(z, x, a) -> F(b(b(f(z), z), x)) C(z, x, a) -> B(b(f(z), z), x) C(z, x, a) -> B(f(z), z) C(z, x, a) -> F(z) B(y, b(z, a)) -> F(b(c(f(a), y, z), z)) B(y, b(z, a)) -> B(c(f(a), y, z), z) B(y, b(z, a)) -> C(f(a), y, z) B(y, b(z, a)) -> F(a) The TRS R consists of the following rules: c(z, x, a) -> f(b(b(f(z), z), x)) b(y, b(z, a)) -> f(b(c(f(a), y, z), z)) f(c(c(z, a, a), x, a)) -> 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 4 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: C(z, x, a) -> B(b(f(z), z), x) B(y, b(z, a)) -> B(c(f(a), y, z), z) B(y, b(z, a)) -> C(f(a), y, z) C(z, x, a) -> B(f(z), z) The TRS R consists of the following rules: c(z, x, a) -> f(b(b(f(z), z), x)) b(y, b(z, a)) -> f(b(c(f(a), y, z), z)) f(c(c(z, a, a), x, a)) -> z Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule C(z, x, a) -> B(b(f(z), z), x) we obtained the following new rules [LPAR04]: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Standard