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TRS Standard pair #516963140
details
property
value
status
complete
benchmark
Ex16_Luc06_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n113.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
3.19369697571 seconds
cpu usage
7.276482826
max memory
5.27699968E8
stage attributes
key
value
output-size
11632
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) MRRProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) TransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) QDP (21) UsableRulesProof [EQUIVALENT, 0 ms] (22) QDP (23) TransformationProof [EQUIVALENT, 0 ms] (24) QDP (25) DependencyGraphProof [EQUIVALENT, 0 ms] (26) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(X, X)) -> mark(f(a, b)) active(b) -> mark(a) mark(f(X1, X2)) -> active(f(mark(X1), X2)) mark(a) -> active(a) mark(b) -> active(b) 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) 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, X)) -> MARK(f(a, b)) ACTIVE(f(X, X)) -> F(a, b) ACTIVE(b) -> MARK(a) MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2)) MARK(f(X1, X2)) -> F(mark(X1), X2) MARK(f(X1, X2)) -> MARK(X1) MARK(a) -> ACTIVE(a) MARK(b) -> ACTIVE(b) 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) The TRS R consists of the following rules: active(f(X, X)) -> mark(f(a, b)) active(b) -> mark(a) mark(f(X1, X2)) -> active(f(mark(X1), X2)) mark(a) -> active(a) mark(b) -> active(b) 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) Q is empty. We have to consider all minimal (P,Q,R)-chains. ----------------------------------------
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