details

property	value
status	complete
benchmark	9.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n073.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.32927 seconds
cpu usage	5.61527
user time	5.33564
system time	0.279624
max virtual memory	1.8477524E7
max residence set size	369484.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPOrderProof [EQUIVALENT, 42 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 30 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x, y) -> b(x, b(0, c(y))) c(b(y, c(x))) -> c(c(b(a(0, 0), y))) b(y, 0) -> 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(x, y) -> B(x, b(0, c(y))) A(x, y) -> B(0, c(y)) A(x, y) -> C(y) C(b(y, c(x))) -> C(c(b(a(0, 0), y))) C(b(y, c(x))) -> C(b(a(0, 0), y)) C(b(y, c(x))) -> B(a(0, 0), y) C(b(y, c(x))) -> A(0, 0) The TRS R consists of the following rules: a(x, y) -> b(x, b(0, c(y))) c(b(y, c(x))) -> c(c(b(a(0, 0), y))) b(y, 0) -> 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(x, y) -> C(y) C(b(y, c(x))) -> C(c(b(a(0, 0), y))) C(b(y, c(x))) -> C(b(a(0, 0), y)) C(b(y, c(x))) -> A(0, 0) The TRS R consists of the following rules: a(x, y) -> b(x, b(0, c(y))) c(b(y, c(x))) -> c(c(b(a(0, 0), y))) b(y, 0) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule A(x, y) -> C(y) we obtained the following new rules [LPAR04]: (A(0, 0) -> C(0),A(0, 0) -> C(0)) ---------------------------------------- (6) Obligation: popout

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

job log

popout

actions

all output return to TRS Standard