details

property	value
status	complete
benchmark	13.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n107.star.cs.uiowa.edu
space	Der95

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	13.1066870689 seconds
cpu usage	26.819074499
max memory	4.594122752E9

stage attributes

key	value
output-size	6283
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 20 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: *(x, +(y, z)) -> +(*(x, y), *(x, z)) *(+(y, z), x) -> +(*(x, y), *(x, z)) *(*(x, y), z) -> *(x, *(y, z)) +(+(x, y), z) -> +(x, +(y, z)) 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: *^1(x, +(y, z)) -> +^1(*(x, y), *(x, z)) *^1(x, +(y, z)) -> *^1(x, y) *^1(x, +(y, z)) -> *^1(x, z) *^1(+(y, z), x) -> +^1(*(x, y), *(x, z)) *^1(+(y, z), x) -> *^1(x, y) *^1(+(y, z), x) -> *^1(x, z) *^1(*(x, y), z) -> *^1(x, *(y, z)) *^1(*(x, y), z) -> *^1(y, z) +^1(+(x, y), z) -> +^1(x, +(y, z)) +^1(+(x, y), z) -> +^1(y, z) The TRS R consists of the following rules: *(x, +(y, z)) -> +(*(x, y), *(x, z)) *(+(y, z), x) -> +(*(x, y), *(x, z)) *(*(x, y), z) -> *(x, *(y, z)) +(+(x, y), z) -> +(x, +(y, z)) 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 2 SCCs with 2 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: +^1(+(x, y), z) -> +^1(y, z) +^1(+(x, y), z) -> +^1(x, +(y, z)) The TRS R consists of the following rules: *(x, +(y, z)) -> +(*(x, y), *(x, z)) *(+(y, z), x) -> +(*(x, y), *(x, z)) *(*(x, y), z) -> *(x, *(y, z)) +(+(x, y), z) -> +(x, +(y, z)) popout

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actions

all output return to TRS Stand 20472