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TRS Stand 20472 pair #381713788
details
property
value
status
complete
benchmark
kabasci03.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n029.star.cs.uiowa.edu
space
AProVE_07
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.52535319328 seconds
cpu usage
7.049993704
max memory
3.99966208E8
stage attributes
key
value
output-size
6271
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 29 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 16 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 81 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: h(c(x, y), c(s(z), z), t(w)) -> h(z, c(y, x), t(t(c(x, c(y, t(w)))))) h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w))) h(c(s(x), c(s(0), y)), z, t(x)) -> h(y, c(s(0), c(x, z)), t(t(c(x, s(x))))) t(t(x)) -> t(c(t(x), x)) t(x) -> x t(x) -> c(0, c(0, c(0, c(0, c(0, 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: H(c(x, y), c(s(z), z), t(w)) -> H(z, c(y, x), t(t(c(x, c(y, t(w)))))) H(c(x, y), c(s(z), z), t(w)) -> T(t(c(x, c(y, t(w))))) H(c(x, y), c(s(z), z), t(w)) -> T(c(x, c(y, t(w)))) H(x, c(y, z), t(w)) -> H(c(s(y), x), z, t(c(t(w), w))) H(x, c(y, z), t(w)) -> T(c(t(w), w)) H(c(s(x), c(s(0), y)), z, t(x)) -> H(y, c(s(0), c(x, z)), t(t(c(x, s(x))))) H(c(s(x), c(s(0), y)), z, t(x)) -> T(t(c(x, s(x)))) H(c(s(x), c(s(0), y)), z, t(x)) -> T(c(x, s(x))) T(t(x)) -> T(c(t(x), x)) The TRS R consists of the following rules: h(c(x, y), c(s(z), z), t(w)) -> h(z, c(y, x), t(t(c(x, c(y, t(w)))))) h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w))) h(c(s(x), c(s(0), y)), z, t(x)) -> h(y, c(s(0), c(x, z)), t(t(c(x, s(x))))) t(t(x)) -> t(c(t(x), x)) t(x) -> x t(x) -> c(0, c(0, c(0, c(0, c(0, 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 6 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: H(x, c(y, z), t(w)) -> H(c(s(y), x), z, t(c(t(w), w))) H(c(x, y), c(s(z), z), t(w)) -> H(z, c(y, x), t(t(c(x, c(y, t(w)))))) H(c(s(x), c(s(0), y)), z, t(x)) -> H(y, c(s(0), c(x, z)), t(t(c(x, s(x))))) The TRS R consists of the following rules: h(c(x, y), c(s(z), z), t(w)) -> h(z, c(y, x), t(t(c(x, c(y, t(w)))))) h(x, c(y, z), t(w)) -> h(c(s(y), x), z, t(c(t(w), w))) h(c(s(x), c(s(0), y)), z, t(x)) -> h(y, c(s(0), c(x, z)), t(t(c(x, s(x))))) t(t(x)) -> t(c(t(x), x)) t(x) -> x t(x) -> c(0, c(0, c(0, c(0, c(0, x))))) Q is empty. We have to consider all minimal (P,Q,R)-chains.
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