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	n172.star.cs.uiowa.edu
space	AProVE_07

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	3.24491882324 seconds
cpu usage	9.418901408
max memory	5.71203584E8

stage attributes

key	value
output-size	6261
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 27 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 13 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 128 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. popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Standard