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	n030.star.cs.uiowa.edu
space	Der95

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	13.9073607922 seconds
cpu usage	32.363991364
max memory	1.689939968E9

stage attributes

key	value
output-size	6276
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPOrderProof [EQUIVALENT, 19 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 Standard