details

property	value
status	complete
benchmark	secr1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	6.16330289841 seconds
cpu usage	21.20079307
max memory	1.324044288E9

stage attributes

key	value
output-size	6735
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 3 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) MRRProof [EQUIVALENT, 0 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 8 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(x1)) -> b(b(b(b(x1)))) b(a(x1)) -> a(a(a(a(x1)))) a(x1) -> x1 b(x1) -> x1 Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(x1)) -> b(b(b(b(x1)))) a(b(x1)) -> a(a(a(a(x1)))) a(x1) -> x1 b(x1) -> x1 Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(x1)) -> B(b(b(b(x1)))) B(a(x1)) -> B(b(b(x1))) B(a(x1)) -> B(b(x1)) B(a(x1)) -> B(x1) A(b(x1)) -> A(a(a(a(x1)))) A(b(x1)) -> A(a(a(x1))) A(b(x1)) -> A(a(x1)) A(b(x1)) -> A(x1) The TRS R consists of the following rules: b(a(x1)) -> b(b(b(b(x1)))) a(b(x1)) -> a(a(a(a(x1)))) a(x1) -> x1 b(x1) -> x1 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) popout

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