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TRS Innermost pair #516979515
details
property
value
status
complete
benchmark
innermost4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n029.star.cs.uiowa.edu
space
Mixed_innermost
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.27828097343 seconds
cpu usage
5.866915498
max memory
4.81247232E8
stage attributes
key
value
output-size
7568
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QReductionProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) TransformationProof [EQUIVALENT, 0 ms] (16) QDP (17) DependencyGraphProof [EQUIVALENT, 0 ms] (18) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a(x), y, s(z), u) -> f(a(b), y, z, g(x, y, s(z), u)) g(x, y, z, u) -> h(x, y, z, u) h(b, y, z, u) -> f(y, y, z, u) a(b) -> c The set Q consists of the following terms: f(a(x0), x1, s(x2), x3) g(x0, x1, x2, x3) h(b, x0, x1, x2) a(b) ---------------------------------------- (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(x), y, s(z), u) -> F(a(b), y, z, g(x, y, s(z), u)) F(a(x), y, s(z), u) -> A(b) F(a(x), y, s(z), u) -> G(x, y, s(z), u) G(x, y, z, u) -> H(x, y, z, u) H(b, y, z, u) -> F(y, y, z, u) The TRS R consists of the following rules: f(a(x), y, s(z), u) -> f(a(b), y, z, g(x, y, s(z), u)) g(x, y, z, u) -> h(x, y, z, u) h(b, y, z, u) -> f(y, y, z, u) a(b) -> c The set Q consists of the following terms: f(a(x0), x1, s(x2), x3) g(x0, x1, x2, x3) h(b, x0, x1, x2) a(b) 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: F(a(x), y, s(z), u) -> G(x, y, s(z), u) G(x, y, z, u) -> H(x, y, z, u) H(b, y, z, u) -> F(y, y, z, u)
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