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TRS Stand 20472 pair #381716057
details
property
value
status
complete
benchmark
Ex1_Luc02b_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n040.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.89418101311 seconds
cpu usage
4.362449375
max memory
2.16195072E8
stage attributes
key
value
output-size
9458
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 6 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) QDPOrderProof [EQUIVALENT, 6 ms] (7) QDP (8) DependencyGraphProof [EQUIVALENT, 0 ms] (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: from(X) -> cons(X, n__from(n__s(X))) first(0, Z) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) sel(0, cons(X, Z)) -> X sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) from(X) -> n__from(X) s(X) -> n__s(X) first(X1, X2) -> n__first(X1, X2) activate(n__from(X)) -> from(activate(X)) activate(n__s(X)) -> s(activate(X)) activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) activate(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: FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z) SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z)) SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z) ACTIVATE(n__from(X)) -> FROM(activate(X)) ACTIVATE(n__from(X)) -> ACTIVATE(X) ACTIVATE(n__s(X)) -> S(activate(X)) ACTIVATE(n__s(X)) -> ACTIVATE(X) ACTIVATE(n__first(X1, X2)) -> FIRST(activate(X1), activate(X2)) ACTIVATE(n__first(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__first(X1, X2)) -> ACTIVATE(X2) The TRS R consists of the following rules: from(X) -> cons(X, n__from(n__s(X))) first(0, Z) -> nil first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) sel(0, cons(X, Z)) -> X sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) from(X) -> n__from(X) s(X) -> n__s(X) first(X1, X2) -> n__first(X1, X2) activate(n__from(X)) -> from(activate(X)) activate(n__s(X)) -> s(activate(X)) activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) activate(X) -> X 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 3 less nodes. ---------------------------------------- (4)
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