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TRS Standard pair #487072798
details
property
value
status
complete
benchmark
4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n181.star.cs.uiowa.edu
space
Secret_06_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.23744 seconds
cpu usage
5.42639
user time
5.18977
system time
0.236621
max virtual memory
1.8482056E7
max residence set size
347024.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) MRRProof [EQUIVALENT, 13 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 27 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(c(b(c(x)))) -> b(a(0, c(x))) c(c(x)) -> b(c(b(c(x)))) a(0, x) -> c(c(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: C(c(b(c(x)))) -> A(0, c(x)) C(c(x)) -> C(b(c(x))) A(0, x) -> C(c(x)) A(0, x) -> C(x) The TRS R consists of the following rules: c(c(b(c(x)))) -> b(a(0, c(x))) c(c(x)) -> b(c(b(c(x)))) a(0, x) -> c(c(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 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(0, x) -> C(c(x)) C(c(b(c(x)))) -> A(0, c(x)) A(0, x) -> C(x) The TRS R consists of the following rules: c(c(b(c(x)))) -> b(a(0, c(x))) c(c(x)) -> b(c(b(c(x)))) a(0, x) -> c(c(x)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: A(0, x) -> C(x) Used ordering: Polynomial interpretation [POLO]: POL(0) = 2 POL(A(x_1, x_2)) = 2*x_1 + 2*x_2 POL(C(x_1)) = 2*x_1 POL(a(x_1, x_2)) = 2*x_1 + x_2 POL(b(x_1)) = x_1 POL(c(x_1)) = 2 + x_1
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