details

property	value
status	complete
benchmark	sigma.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	Mixed_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.25372 seconds
cpu usage	8.1085
user time	7.79672
system time	0.311775
max virtual memory	1.8411992E7
max residence set size	476520.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPOrderProof [EQUIVALENT, 817 ms] (6) QDP (7) QDPSizeChangeProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: comp(s, id) -> s cons(one, shift) -> id cons(comp(one, s), comp(shift, s)) -> s comp(one, cons(s, t)) -> s comp(shift, cons(s, t)) -> t comp(abs(s), t) -> abs(comp(s, cons(one, comp(t, shift)))) comp(cons(s, t), u) -> cons(comp(s, u), comp(t, u)) comp(id, s) -> s comp(comp(s, t), u) -> comp(s, comp(t, u)) 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: COMP(abs(s), t) -> COMP(s, cons(one, comp(t, shift))) COMP(abs(s), t) -> CONS(one, comp(t, shift)) COMP(abs(s), t) -> COMP(t, shift) COMP(cons(s, t), u) -> CONS(comp(s, u), comp(t, u)) COMP(cons(s, t), u) -> COMP(s, u) COMP(cons(s, t), u) -> COMP(t, u) COMP(comp(s, t), u) -> COMP(s, comp(t, u)) COMP(comp(s, t), u) -> COMP(t, u) The TRS R consists of the following rules: comp(s, id) -> s cons(one, shift) -> id cons(comp(one, s), comp(shift, s)) -> s comp(one, cons(s, t)) -> s comp(shift, cons(s, t)) -> t comp(abs(s), t) -> abs(comp(s, cons(one, comp(t, shift)))) comp(cons(s, t), u) -> cons(comp(s, u), comp(t, u)) comp(id, s) -> s comp(comp(s, t), u) -> comp(s, comp(t, u)) 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 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: COMP(abs(s), t) -> COMP(t, shift) COMP(abs(s), t) -> COMP(s, cons(one, comp(t, shift))) COMP(cons(s, t), u) -> COMP(s, u) COMP(cons(s, t), u) -> COMP(t, u) COMP(comp(s, t), u) -> COMP(s, comp(t, u)) COMP(comp(s, t), u) -> COMP(t, u) The TRS R consists of the following rules: comp(s, id) -> s cons(one, shift) -> id cons(comp(one, s), comp(shift, s)) -> s comp(one, cons(s, t)) -> s comp(shift, cons(s, t)) -> t comp(abs(s), t) -> abs(comp(s, cons(one, comp(t, shift)))) comp(cons(s, t), u) -> cons(comp(s, u), comp(t, u)) comp(id, s) -> s comp(comp(s, t), u) -> comp(s, comp(t, u)) popout

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all output return to TRS Standard