details

property	value
status	complete
benchmark	ttt1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n179.star.cs.uiowa.edu
space	Secret_05_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.43793 seconds
cpu usage	5.51905
user time	5.2685
system time	0.250549
max virtual memory	3.6972948E7
max residence set size	320244.0

stage attributes

key	value
starexec-result	NO

output

NO 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) NonTerminationLoopProof [COMPLETE, 0 ms] (7) NO (8) QDP (9) NonTerminationLoopProof [COMPLETE, 26 ms] (10) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(a), s(b), x) -> f(x, x, x) g(f(s(x), s(y), z)) -> g(f(x, y, z)) cons(x, y) -> x cons(x, y) -> y 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: F(s(a), s(b), x) -> F(x, x, x) G(f(s(x), s(y), z)) -> G(f(x, y, z)) G(f(s(x), s(y), z)) -> F(x, y, z) The TRS R consists of the following rules: f(s(a), s(b), x) -> f(x, x, x) g(f(s(x), s(y), z)) -> g(f(x, y, z)) cons(x, y) -> x cons(x, y) -> y 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 2 SCCs with 1 less node. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(a), s(b), x) -> F(x, x, x) The TRS R consists of the following rules: f(s(a), s(b), x) -> f(x, x, x) g(f(s(x), s(y), z)) -> g(f(x, y, z)) cons(x, y) -> x cons(x, y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = F(cons(s(a), y), cons(x', s(b)), x) evaluates to t =F(x, x, x) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [y / s(b), x' / s(a), x / cons(s(a), s(b))] -------------------------------------------------------------------------------- Rewriting sequence popout

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job log

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actions

all output return to TRS Standard