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TRS Standard pair #487069818
details
property
value
status
complete
benchmark
Ex6_Luc98_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n070.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.13911 seconds
cpu usage
5.2246
user time
4.99741
system time
0.227196
max virtual memory
1.8411988E7
max residence set size
361940.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 14 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 51 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__first(0, X) -> nil a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) a__from(X) -> cons(mark(X), from(s(X))) mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) mark(from(X)) -> a__from(mark(X)) mark(0) -> 0 mark(nil) -> nil mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__first(X1, X2) -> first(X1, X2) a__from(X) -> from(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: A__FIRST(s(X), cons(Y, Z)) -> MARK(Y) A__FROM(X) -> MARK(X) MARK(first(X1, X2)) -> A__FIRST(mark(X1), mark(X2)) MARK(first(X1, X2)) -> MARK(X1) MARK(first(X1, X2)) -> MARK(X2) MARK(from(X)) -> A__FROM(mark(X)) MARK(from(X)) -> MARK(X) MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: a__first(0, X) -> nil a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) a__from(X) -> cons(mark(X), from(s(X))) mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) mark(from(X)) -> a__from(mark(X)) mark(0) -> 0 mark(nil) -> nil mark(s(X)) -> s(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) a__first(X1, X2) -> first(X1, X2) a__from(X) -> from(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A__FIRST(s(X), cons(Y, Z)) -> MARK(Y) MARK(from(X)) -> A__FROM(mark(X)) MARK(from(X)) -> MARK(X) MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A__FIRST_2(x_1, x_2) ) = max{0, x_1 + x_2 - 1} POL( A__FROM_1(x_1) ) = x_1 POL( mark_1(x_1) ) = 2x_1 POL( first_2(x_1, x_2) ) = 2x_1 + 2x_2 POL( a__first_2(x_1, x_2) ) = 2x_1 + 2x_2 POL( from_1(x_1) ) = 2x_1 + 2
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