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TRS Stand 20472 pair #381715737
details
property
value
status
complete
benchmark
Ex9_BLR02_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n080.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.11292600632 seconds
cpu usage
5.203294058
max memory
3.49925376E8
stage attributes
key
value
output-size
9190
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPOrderProof [EQUIVALENT, 57 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: filter(cons(X, Y), 0, M) -> cons(0, n__filter(activate(Y), M, M)) filter(cons(X, Y), s(N), M) -> cons(X, n__filter(activate(Y), N, M)) sieve(cons(0, Y)) -> cons(0, n__sieve(activate(Y))) sieve(cons(s(N), Y)) -> cons(s(N), n__sieve(filter(activate(Y), N, N))) nats(N) -> cons(N, n__nats(s(N))) zprimes -> sieve(nats(s(s(0)))) filter(X1, X2, X3) -> n__filter(X1, X2, X3) sieve(X) -> n__sieve(X) nats(X) -> n__nats(X) activate(n__filter(X1, X2, X3)) -> filter(X1, X2, X3) activate(n__sieve(X)) -> sieve(X) activate(n__nats(X)) -> nats(X) 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: FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y) FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y) SIEVE(cons(0, Y)) -> ACTIVATE(Y) SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N) SIEVE(cons(s(N), Y)) -> ACTIVATE(Y) ZPRIMES -> SIEVE(nats(s(s(0)))) ZPRIMES -> NATS(s(s(0))) ACTIVATE(n__filter(X1, X2, X3)) -> FILTER(X1, X2, X3) ACTIVATE(n__sieve(X)) -> SIEVE(X) ACTIVATE(n__nats(X)) -> NATS(X) The TRS R consists of the following rules: filter(cons(X, Y), 0, M) -> cons(0, n__filter(activate(Y), M, M)) filter(cons(X, Y), s(N), M) -> cons(X, n__filter(activate(Y), N, M)) sieve(cons(0, Y)) -> cons(0, n__sieve(activate(Y))) sieve(cons(s(N), Y)) -> cons(s(N), n__sieve(filter(activate(Y), N, N))) nats(N) -> cons(N, n__nats(s(N))) zprimes -> sieve(nats(s(s(0)))) filter(X1, X2, X3) -> n__filter(X1, X2, X3) sieve(X) -> n__sieve(X) nats(X) -> n__nats(X) activate(n__filter(X1, X2, X3)) -> filter(X1, X2, X3) activate(n__sieve(X)) -> sieve(X) activate(n__nats(X)) -> nats(X) 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 1 SCC with 3 less nodes. ---------------------------------------- (4) Obligation: Q DP problem:
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