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TRS Stand 20472 pair #381709621
details
property
value
status
complete
benchmark
Liveness_WRS.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n016.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.98308396339 seconds
cpu usage
4.94742911
max memory
2.89468416E8
stage attributes
key
value
output-size
12468
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) MRRProof [EQUIVALENT, 0 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) MNOCProof [EQUIVALENT, 0 ms] (16) QDP (17) MRRProof [EQUIVALENT, 0 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) TRUE (21) QDP (22) UsableRulesProof [EQUIVALENT, 0 ms] (23) QDP (24) TransformationProof [EQUIVALENT, 0 ms] (25) QDP (26) DependencyGraphProof [EQUIVALENT, 0 ms] (27) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(x)) -> mark(f(f(x))) chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))) mat(f(x), f(y)) -> f(mat(x, y)) chk(no(c)) -> active(c) mat(f(x), c) -> no(c) f(active(x)) -> active(f(x)) f(no(x)) -> no(f(x)) f(mark(x)) -> mark(f(x)) tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), 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(x)) -> F(f(x)) CHK(no(f(x))) -> F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))) CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)) CHK(no(f(x))) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x) CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(f(X)))))))))) CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(X))))))))) CHK(no(f(x))) -> F(f(f(f(f(f(f(f(X)))))))) CHK(no(f(x))) -> F(f(f(f(f(f(f(X))))))) CHK(no(f(x))) -> F(f(f(f(f(f(X)))))) CHK(no(f(x))) -> F(f(f(f(f(X))))) CHK(no(f(x))) -> F(f(f(f(X)))) CHK(no(f(x))) -> F(f(f(X))) CHK(no(f(x))) -> F(f(X)) CHK(no(f(x))) -> F(X) MAT(f(x), f(y)) -> F(mat(x, y)) MAT(f(x), f(y)) -> MAT(x, y) CHK(no(c)) -> ACTIVE(c) F(active(x)) -> ACTIVE(f(x)) F(active(x)) -> F(x) F(no(x)) -> F(x) F(mark(x)) -> F(x) TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))) TP(mark(x)) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)) TP(mark(x)) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x) TP(mark(x)) -> F(f(f(f(f(f(f(f(f(f(X)))))))))) TP(mark(x)) -> F(f(f(f(f(f(f(f(f(X))))))))) TP(mark(x)) -> F(f(f(f(f(f(f(f(X))))))))
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