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TRS Standard pair #516961075
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
n138.star.cs.uiowa.edu
space
Secret_05_TRS
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.5706319809 seconds
cpu usage
7.023185154
max memory
5.1161088E8
stage attributes
key
value
output-size
5439
starexec-result
NO
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 5 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) NonTerminationLoopProof [COMPLETE, 0 ms] (7) NO (8) QDP (9) NonTerminationLoopProof [COMPLETE, 24 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:
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