details

property	value
status	complete
benchmark	jones2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n087.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.14091205597 seconds
cpu usage	5.618728448
max memory	4.80534528E8

stage attributes

key	value
output-size	3783
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 14 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(empty, l) -> l f(cons(x, k), l) -> g(k, l, cons(x, k)) g(a, b, c) -> f(a, cons(b, c)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(empty, l) -> l f(cons(x, k), l) -> g(k, l, cons(x, k)) g(a, b, c) -> f(a, cons(b, c)) The set Q consists of the following terms: f(empty, x0) f(cons(x0, x1), x2) g(x0, 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: F(cons(x, k), l) -> G(k, l, cons(x, k)) G(a, b, c) -> F(a, cons(b, c)) The TRS R consists of the following rules: f(empty, l) -> l f(cons(x, k), l) -> g(k, l, cons(x, k)) g(a, b, c) -> f(a, cons(b, c)) The set Q consists of the following terms: f(empty, x0) f(cons(x0, x1), x2) g(x0, 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: The TRS P consists of the following rules: F(cons(x, k), l) -> G(k, l, cons(x, k)) popout

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