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TRS Standard pair #516967500
details
property
value
status
complete
benchmark
2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n114.star.cs.uiowa.edu
space
Secret_07_TRS
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
35.1774039268 seconds
cpu usage
115.775402654
max memory
3.91165952E9
stage attributes
key
value
output-size
13840
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 4 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPOrderProof [EQUIVALENT, 19 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) TransformationProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPBoundsTAProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPBoundsTAProof [EQUIVALENT, 6 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) QDPOrderProof [EQUIVALENT, 60 ms] (25) QDP (26) TransformationProof [EQUIVALENT, 0 ms] (27) QDP (28) QDPOrderProof [EQUIVALENT, 2146 ms] (29) QDP (30) PisEmptyProof [EQUIVALENT, 0 ms] (31) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(x, x) -> h(a, b) g(g(x, a), y) -> g(g(a, y), g(a, x)) f(g(x, y)) -> g(g(f(f(y)), h(a, a)), x) h(h(f(f(x)), y), h(z, v)) -> h(h(f(z), f(f(f(y)))), h(v, x)) Q is empty. ---------------------------------------- (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: H(x, x) -> H(a, b) G(g(x, a), y) -> G(g(a, y), g(a, x)) G(g(x, a), y) -> G(a, y) G(g(x, a), y) -> G(a, x) F(g(x, y)) -> G(g(f(f(y)), h(a, a)), x) F(g(x, y)) -> G(f(f(y)), h(a, a)) F(g(x, y)) -> F(f(y)) F(g(x, y)) -> F(y) F(g(x, y)) -> H(a, a) H(h(f(f(x)), y), h(z, v)) -> H(h(f(z), f(f(f(y)))), h(v, x)) H(h(f(f(x)), y), h(z, v)) -> H(f(z), f(f(f(y)))) H(h(f(f(x)), y), h(z, v)) -> F(z) H(h(f(f(x)), y), h(z, v)) -> F(f(f(y))) H(h(f(f(x)), y), h(z, v)) -> F(f(y)) H(h(f(f(x)), y), h(z, v)) -> F(y) H(h(f(f(x)), y), h(z, v)) -> H(v, x) The TRS R consists of the following rules: h(x, x) -> h(a, b) g(g(x, a), y) -> g(g(a, y), g(a, x)) f(g(x, y)) -> g(g(f(f(y)), h(a, a)), x) h(h(f(f(x)), y), h(z, v)) -> h(h(f(z), f(f(f(y)))), h(v, x)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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