details

property	value
status	complete
benchmark	direct.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n111.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.05352592468 seconds
cpu usage	4.86973138
max memory	2.681856E8

stage attributes

key	value
output-size	4736
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 6 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(x, c(y, z)) -> h(c(s(y), x), z) h(c(s(x), c(s(0), y)), z) -> h(y, c(s(0), c(x, z))) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none The TRS R 2 is h(x, c(y, z)) -> h(c(s(y), x), z) h(c(s(x), c(s(0), y)), z) -> h(y, c(s(0), c(x, z))) The signature Sigma is {h_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(x, c(y, z)) -> h(c(s(y), x), z) h(c(s(x), c(s(0), y)), z) -> h(y, c(s(0), c(x, z))) The set Q consists of the following terms: h(x0, c(x1, x2)) h(c(s(x0), c(s(0), x1)), x2) ---------------------------------------- (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: H(x, c(y, z)) -> H(c(s(y), x), z) H(c(s(x), c(s(0), y)), z) -> H(y, c(s(0), c(x, z))) The TRS R consists of the following rules: h(x, c(y, z)) -> h(c(s(y), x), z) h(c(s(x), c(s(0), y)), z) -> h(y, c(s(0), c(x, z))) The set Q consists of the following terms: h(x0, c(x1, x2)) h(c(s(x0), c(s(0), x1)), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: popout

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

job log

popout

actions

all output return to TRS Stand 20472