details

property	value
status	complete
benchmark	10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n179.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.41815 seconds
cpu usage	9.44557
user time	9.02812
system time	0.41745
max virtual memory	1.8617272E7
max residence set size	599372.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 25 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 235 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(c(c(y))) -> c(c(a(y, 0))) c(a(a(0, x), y)) -> a(c(c(c(0))), y) c(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: C(c(c(y))) -> C(c(a(y, 0))) C(c(c(y))) -> C(a(y, 0)) C(a(a(0, x), y)) -> C(c(c(0))) C(a(a(0, x), y)) -> C(c(0)) C(a(a(0, x), y)) -> C(0) The TRS R consists of the following rules: c(c(c(y))) -> c(c(a(y, 0))) c(a(a(0, x), y)) -> a(c(c(c(0))), y) c(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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: C(c(c(y))) -> C(a(y, 0)) C(a(a(0, x), y)) -> C(c(c(0))) C(c(c(y))) -> C(c(a(y, 0))) C(a(a(0, x), y)) -> C(c(0)) The TRS R consists of the following rules: c(c(c(y))) -> c(c(a(y, 0))) c(a(a(0, x), y)) -> a(c(c(c(0))), y) c(y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule C(a(a(0, x), y)) -> C(c(0)) at position [0] we obtained the following new rules [LPAR04]: (C(a(a(0, y0), y1)) -> C(0),C(a(a(0, y0), y1)) -> C(0)) ---------------------------------------- (6) Obligation: popout

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

job log

popout

actions

all output return to TRS Standard