details

property	value
status	complete
benchmark	289.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n048.star.cs.uiowa.edu
space	COPS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.56647396088 seconds
cpu usage	6.49336316
max memory	4.02378752E8

stage attributes

key	value
output-size	4954
starexec-result	MAYBE

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Quasi decreasingness of the given CTRS could not be shown: (0) CTRS (1) CTRSToQTRSProof [SOUND, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) NonTerminationLoopProof [COMPLETE, 270 ms] (8) NO ---------------------------------------- (0) Obligation: Conditional term rewrite system: The TRS R consists of the following rules: App(App(App(S, x), y), z) -> App(App(x, z), App(y, z)) App(App(K, x), y) -> x App(I, x) -> x App(App(App(C, T), x), y) -> x App(App(App(C, F), x), y) -> y The conditional TRS C consists of the following conditional rules: App(App(App(C, z), x), y) -> x <= x -> y App(App(App(C, z), x), y) -> y <= x -> y ---------------------------------------- (1) CTRSToQTRSProof (SOUND) The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: App(App(App(C, z), x), y) -> U1(x, x, y) U1(y, x, y) -> y U1(y, x, y) -> x App(App(App(S, x), y), z) -> App(App(x, z), App(y, z)) App(App(K, x), y) -> x App(I, x) -> x App(App(App(C, T), x), y) -> x App(App(App(C, F), x), y) -> y Q is empty. ---------------------------------------- (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: APP(App(App(C, z), x), y) -> U1^1(x, x, y) APP(App(App(S, x), y), z) -> APP(App(x, z), App(y, z)) APP(App(App(S, x), y), z) -> APP(x, z) APP(App(App(S, x), y), z) -> APP(y, z) The TRS R consists of the following rules: App(App(App(C, z), x), y) -> U1(x, x, y) U1(y, x, y) -> y U1(y, x, y) -> x App(App(App(S, x), y), z) -> App(App(x, z), App(y, z)) App(App(K, x), y) -> x App(I, x) -> x App(App(App(C, T), x), y) -> x App(App(App(C, F), x), y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- popout

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actions

all output return to TRS Condi 20667