details

property	value
status	complete
benchmark	1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n035.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.36138486862 seconds
cpu usage	6.335956595
max memory	4.83278848E8

stage attributes

key	value
output-size	3486
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 40 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, f(a, x)) -> f(a, f(x, f(f(a, a), a))) 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: F(a, f(a, x)) -> F(a, f(x, f(f(a, a), a))) F(a, f(a, x)) -> F(x, f(f(a, a), a)) F(a, f(a, x)) -> F(f(a, a), a) F(a, f(a, x)) -> F(a, a) The TRS R consists of the following rules: f(a, f(a, x)) -> f(a, f(x, f(f(a, a), a))) 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 1 SCC with 3 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(a, f(a, x)) -> F(a, f(x, f(f(a, a), a))) The TRS R consists of the following rules: f(a, f(a, x)) -> f(a, f(x, f(f(a, a), a))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(a, f(a, x)) -> F(a, f(x, f(f(a, a), a))) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. popout

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job log

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actions

all output return to TRS Standard