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TRS Standard pair #516963350
details
property
value
status
complete
benchmark
Ex9_Luc06_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n115.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.66708612442 seconds
cpu usage
7.247187571
max memory
5.026816E8
stage attributes
key
value
output-size
9766
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 13 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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, 29 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 74 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(a, X, X)) -> mark(f(X, b, b)) active(b) -> mark(a) mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3)) mark(a) -> active(a) mark(b) -> active(b) f(mark(X1), X2, X3) -> f(X1, X2, X3) f(X1, mark(X2), X3) -> f(X1, X2, X3) f(X1, X2, mark(X3)) -> f(X1, X2, X3) f(active(X1), X2, X3) -> f(X1, X2, X3) f(X1, active(X2), X3) -> f(X1, X2, X3) f(X1, X2, active(X3)) -> f(X1, X2, X3) 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(a, X, X)) -> MARK(f(X, b, b)) ACTIVE(f(a, X, X)) -> F(X, b, b) ACTIVE(b) -> MARK(a) MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, mark(X2), X3)) MARK(f(X1, X2, X3)) -> F(X1, mark(X2), X3) MARK(f(X1, X2, X3)) -> MARK(X2) MARK(a) -> ACTIVE(a) MARK(b) -> ACTIVE(b) F(mark(X1), X2, X3) -> F(X1, X2, X3) F(X1, mark(X2), X3) -> F(X1, X2, X3) F(X1, X2, mark(X3)) -> F(X1, X2, X3) F(active(X1), X2, X3) -> F(X1, X2, X3) F(X1, active(X2), X3) -> F(X1, X2, X3) F(X1, X2, active(X3)) -> F(X1, X2, X3) The TRS R consists of the following rules: active(f(a, X, X)) -> mark(f(X, b, b)) active(b) -> mark(a) mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3)) mark(a) -> active(a) mark(b) -> active(b) f(mark(X1), X2, X3) -> f(X1, X2, X3) f(X1, mark(X2), X3) -> f(X1, X2, X3) f(X1, X2, mark(X3)) -> f(X1, X2, X3) f(active(X1), X2, X3) -> f(X1, X2, X3) f(X1, active(X2), X3) -> f(X1, X2, X3) f(X1, X2, active(X3)) -> f(X1, X2, X3) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes. ----------------------------------------
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