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TRS Stand 20472 pair #381712194
details
property
value
status
complete
benchmark
secret5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n045.star.cs.uiowa.edu
space
Secret_07_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.01474499702 seconds
cpu usage
4.896705455
max memory
3.43416832E8
stage attributes
key
value
output-size
12662
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) UsableRulesProof [EQUIVALENT, 0 ms] (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES (20) QDP (21) QDPOrderProof [EQUIVALENT, 37 ms] (22) QDP (23) DependencyGraphProof [EQUIVALENT, 0 ms] (24) QDP (25) UsableRulesProof [EQUIVALENT, 0 ms] (26) QDP (27) QDPSizeChangeProof [EQUIVALENT, 0 ms] (28) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: t(N) -> cs(r(q(N)), nt(ns(N))) q(0) -> 0 q(s(X)) -> s(p(q(X), d(X))) d(0) -> 0 d(s(X)) -> s(s(d(X))) p(0, X) -> X p(X, 0) -> X p(s(X), s(Y)) -> s(s(p(X, Y))) f(0, X) -> nil f(s(X), cs(Y, Z)) -> cs(Y, nf(X, a(Z))) t(X) -> nt(X) s(X) -> ns(X) f(X1, X2) -> nf(X1, X2) a(nt(X)) -> t(a(X)) a(ns(X)) -> s(a(X)) a(nf(X1, X2)) -> f(a(X1), a(X2)) a(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: T(N) -> Q(N) Q(s(X)) -> S(p(q(X), d(X))) Q(s(X)) -> P(q(X), d(X)) Q(s(X)) -> Q(X) Q(s(X)) -> D(X) D(s(X)) -> S(s(d(X))) D(s(X)) -> S(d(X)) D(s(X)) -> D(X) P(s(X), s(Y)) -> S(s(p(X, Y))) P(s(X), s(Y)) -> S(p(X, Y)) P(s(X), s(Y)) -> P(X, Y) F(s(X), cs(Y, Z)) -> A(Z) A(nt(X)) -> T(a(X)) A(nt(X)) -> A(X) A(ns(X)) -> S(a(X)) A(ns(X)) -> A(X) A(nf(X1, X2)) -> F(a(X1), a(X2)) A(nf(X1, X2)) -> A(X1)
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