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TRS Stand 20472 pair #381710970
details
property
value
status
complete
benchmark
t013.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n109.star.cs.uiowa.edu
space
HirokawaMiddeldorp_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.78988289833 seconds
cpu usage
4.144431712
max memory
2.54988288E8
stage attributes
key
value
output-size
7242
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPOrderProof [EQUIVALENT, 25 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: -(x, 0) -> x -(0, s(y)) -> 0 -(s(x), s(y)) -> -(x, y) f(0) -> 0 f(s(x)) -> -(s(x), g(f(x))) g(0) -> s(0) g(s(x)) -> -(s(x), f(g(x))) 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: -(x, 0) -> x -(0, s(y)) -> 0 -(s(x), s(y)) -> -(x, y) f(0) -> 0 f(s(x)) -> -(s(x), g(f(x))) g(0) -> s(0) g(s(x)) -> -(s(x), f(g(x))) The set Q consists of the following terms: -(x0, 0) -(0, s(x0)) -(s(x0), s(x1)) f(0) f(s(x0)) g(0) g(s(x0)) ---------------------------------------- (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: -^1(s(x), s(y)) -> -^1(x, y) F(s(x)) -> -^1(s(x), g(f(x))) F(s(x)) -> G(f(x)) F(s(x)) -> F(x) G(s(x)) -> -^1(s(x), f(g(x))) G(s(x)) -> F(g(x)) G(s(x)) -> G(x)
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