details

property	value
status	complete
benchmark	#4.23.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n030.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.31176018715 seconds
cpu usage	6.231143486
max memory	4.79346688E8

stage attributes

key	value
output-size	9111
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 3 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: quot(0, s(y), s(z)) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z))) The set Q consists of the following terms: quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) plus(0, x0) plus(s(x0), x1) quot(x0, 0, s(x1)) ---------------------------------------- (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: QUOT(s(x), s(y), z) -> QUOT(x, y, z) PLUS(s(x), y) -> PLUS(x, y) QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z)) QUOT(x, 0, s(z)) -> PLUS(z, s(0)) The TRS R consists of the following rules: quot(0, s(y), s(z)) -> 0 quot(s(x), s(y), z) -> quot(x, y, z) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z))) The set Q consists of the following terms: quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) plus(0, x0) plus(s(x0), x1) quot(x0, 0, s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (4) popout

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

job log

popout

actions

all output return to TRS Innermost