details

property	value
status	complete
benchmark	matchbox1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n113.star.cs.uiowa.edu
space	Secret_05_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.07031583786 seconds
cpu usage	4.988900692
max memory	2.75509248E8

stage attributes

key	value
output-size	4465
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) TransformationProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 5 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(f(x, a), y) -> f(y, f(x, 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: F(f(x, a), y) -> F(y, f(x, y)) F(f(x, a), y) -> F(x, y) The TRS R consists of the following rules: f(f(x, a), y) -> f(y, f(x, y)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule F(f(x, a), y) -> F(y, f(x, y)) we obtained the following new rules [LPAR04]: (F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a))),F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a)))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(f(x, a), y) -> F(x, y) F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a))) The TRS R consists of the following rules: f(f(x, a), y) -> f(y, f(x, y)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule F(f(x, a), y) -> F(x, y) we obtained the following new rules [LPAR04]: (F(f(f(y_0, a), a), x1) -> F(f(y_0, a), x1),F(f(f(y_0, a), a), x1) -> F(f(y_0, a), x1)) (F(f(f(y_0, a), a), f(y_1, a)) -> F(f(y_0, a), f(y_1, a)),F(f(f(y_0, a), a), f(y_1, a)) -> F(f(y_0, a), f(y_1, a))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(f(x0, a), f(y_0, a)) -> F(f(y_0, a), f(x0, f(y_0, a))) F(f(f(y_0, a), a), x1) -> F(f(y_0, a), x1) F(f(f(y_0, a), a), f(y_1, a)) -> F(f(y_0, a), f(y_1, a)) popout

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job log

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actions

all output return to TRS Stand 20472