Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381712198
details
property
value
status
complete
benchmark
jones2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n106.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.64588093758 seconds
cpu usage
3.530087425
max memory
2.24534528E8
stage attributes
key
value
output-size
3789
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(empty, l) -> l f(cons(x, k), l) -> g(k, l, cons(x, k)) g(a, b, c) -> f(a, cons(b, c)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(empty, l) -> l f(cons(x, k), l) -> g(k, l, cons(x, k)) g(a, b, c) -> f(a, cons(b, c)) The set Q consists of the following terms: f(empty, x0) f(cons(x0, x1), x2) g(x0, x1, x2) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(cons(x, k), l) -> G(k, l, cons(x, k)) G(a, b, c) -> F(a, cons(b, c)) The TRS R consists of the following rules: f(empty, l) -> l f(cons(x, k), l) -> g(k, l, cons(x, k)) g(a, b, c) -> f(a, cons(b, c)) The set Q consists of the following terms: f(empty, x0) f(cons(x0, x1), x2) g(x0, x1, x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(cons(x, k), l) -> G(k, l, cons(x, k))
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472