details

property	value
status	complete
benchmark	#4.2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n082.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.08031201363 seconds
cpu usage	5.530596263
max memory	4.81329152E8

stage attributes

key	value
output-size	3570
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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) 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(g(x), s(0), y) -> f(y, y, g(x)) g(s(x)) -> s(g(x)) g(0) -> 0 The set Q consists of the following terms: f(g(x0), s(0), x1) g(s(x0)) g(0) ---------------------------------------- (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(g(x), s(0), y) -> F(y, y, g(x)) G(s(x)) -> G(x) The TRS R consists of the following rules: f(g(x), s(0), y) -> f(y, y, g(x)) g(s(x)) -> s(g(x)) g(0) -> 0 The set Q consists of the following terms: f(g(x0), s(0), x1) g(s(x0)) g(0) 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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: G(s(x)) -> G(x) The TRS R consists of the following rules: f(g(x), s(0), y) -> f(y, y, g(x)) g(s(x)) -> s(g(x)) g(0) -> 0 The set Q consists of the following terms: f(g(x0), s(0), x1) g(s(x0)) g(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) popout

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job log

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actions

all output return to TRS Innermost