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TRS Stand 20472 pair #381716062
details
property
value
status
complete
benchmark
Ex4_4_Luc96b_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n102.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.79917001724 seconds
cpu usage
4.095530313
max memory
2.58777088E8
stage attributes
key
value
output-size
10781
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 29 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) QDPOrderProof [EQUIVALENT, 12 ms] (17) QDP (18) DependencyGraphProof [EQUIVALENT, 0 ms] (19) QDP (20) UsableRulesProof [EQUIVALENT, 0 ms] (21) QDP (22) QDPSizeChangeProof [EQUIVALENT, 0 ms] (23) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(g(X), Y)) -> mark(f(X, f(g(X), Y))) mark(f(X1, X2)) -> active(f(mark(X1), X2)) mark(g(X)) -> active(g(mark(X))) f(mark(X1), X2) -> f(X1, X2) f(X1, mark(X2)) -> f(X1, X2) f(active(X1), X2) -> f(X1, X2) f(X1, active(X2)) -> f(X1, X2) g(mark(X)) -> g(X) g(active(X)) -> g(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: ACTIVE(f(g(X), Y)) -> MARK(f(X, f(g(X), Y))) ACTIVE(f(g(X), Y)) -> F(X, f(g(X), Y)) MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2)) MARK(f(X1, X2)) -> F(mark(X1), X2) MARK(f(X1, X2)) -> MARK(X1) MARK(g(X)) -> ACTIVE(g(mark(X))) MARK(g(X)) -> G(mark(X)) MARK(g(X)) -> MARK(X) F(mark(X1), X2) -> F(X1, X2) F(X1, mark(X2)) -> F(X1, X2) F(active(X1), X2) -> F(X1, X2) F(X1, active(X2)) -> F(X1, X2) G(mark(X)) -> G(X) G(active(X)) -> G(X) The TRS R consists of the following rules: active(f(g(X), Y)) -> mark(f(X, f(g(X), Y))) mark(f(X1, X2)) -> active(f(mark(X1), X2)) mark(g(X)) -> active(g(mark(X))) f(mark(X1), X2) -> f(X1, X2) f(X1, mark(X2)) -> f(X1, X2) f(active(X1), X2) -> f(X1, X2) f(X1, active(X2)) -> f(X1, X2) g(mark(X)) -> g(X) g(active(X)) -> g(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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