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TRS Conditional pair #516978671
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
n072.star.cs.uiowa.edu
space
COPS
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
3.25023913383 seconds
cpu usage
9.172999227
max memory
6.27195904E8
stage attributes
key
value
output-size
4952
starexec-result
MAYBE
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Quasi decreasingness of the given CTRS could not be shown: (0) CTRS (1) CTRSToQTRSProof [SOUND, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 18 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) NonTerminationLoopProof [COMPLETE, 339 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. ----------------------------------------
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