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	n173.star.cs.uiowa.edu
space	Der95

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	13.1879 seconds
cpu usage	28.0523
user time	26.4131
system time	1.63919
max virtual memory	1.9139164E7
max residence set size	4486424.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 2 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 17 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)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) popout

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