details

property	value
status	complete
benchmark	#3.40.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n054.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.5837469101 seconds
cpu usage	7.093205985
max memory	5.24115968E8

stage attributes

key	value
output-size	12179
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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) MRRProof [EQUIVALENT, 22 ms] (9) QDP (10) MRRProof [EQUIVALENT, 21 ms] (11) QDP (12) MRRProof [EQUIVALENT, 22 ms] (13) QDP (14) DependencyGraphProof [EQUIVALENT, 0 ms] (15) TRUE (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES (21) QDP (22) QDPOrderProof [EQUIVALENT, 5 ms] (23) QDP (24) PisEmptyProof [EQUIVALENT, 0 ms] (25) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0))) plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0))) 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: MINUS(s(x), s(y)) -> MINUS(x, y) QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y)) QUOT(s(x), s(y)) -> MINUS(x, y) PLUS(s(x), y) -> PLUS(x, y) PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0))) PLUS(plus(x, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), plus(x, s(0))) The TRS R consists of the following rules: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) plus(0, y) -> y plus(s(x), y) -> s(plus(x, y)) plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0))) plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(x, s(0))) 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 3 SCCs with 1 less node. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- popout

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

job log

popout

actions

all output return to TRS Standard