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TRS Stand 20472 pair #381711904
details
property
value
status
complete
benchmark
logarquot.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n026.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
muterm 5.18
configuration
default
runtime (wallclock)
0.0494871139526 seconds
cpu usage
0.047892388
max memory
3715072.0
stage attributes
key
value
output-size
4719
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR X Y) (RULES log(s(0)) -> 0 log(s(s(X))) -> s(log(s(quot(X,s(s(0)))))) min(s(X),s(Y)) -> min(X,Y) min(X,0) -> X quot(0,s(Y)) -> 0 quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y))) ) Problem 1: Innermost Equivalent Processor: -> Rules: log(s(0)) -> 0 log(s(s(X))) -> s(log(s(quot(X,s(s(0)))))) min(s(X),s(Y)) -> min(X,Y) min(X,0) -> X quot(0,s(Y)) -> 0 quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y))) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: LOG(s(s(X))) -> LOG(s(quot(X,s(s(0))))) LOG(s(s(X))) -> QUOT(X,s(s(0))) MIN(s(X),s(Y)) -> MIN(X,Y) QUOT(s(X),s(Y)) -> MIN(X,Y) QUOT(s(X),s(Y)) -> QUOT(min(X,Y),s(Y)) -> Rules: log(s(0)) -> 0 log(s(s(X))) -> s(log(s(quot(X,s(s(0)))))) min(s(X),s(Y)) -> min(X,Y) min(X,0) -> X quot(0,s(Y)) -> 0 quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y))) Problem 1: SCC Processor: -> Pairs: LOG(s(s(X))) -> LOG(s(quot(X,s(s(0))))) LOG(s(s(X))) -> QUOT(X,s(s(0))) MIN(s(X),s(Y)) -> MIN(X,Y) QUOT(s(X),s(Y)) -> MIN(X,Y) QUOT(s(X),s(Y)) -> QUOT(min(X,Y),s(Y)) -> Rules: log(s(0)) -> 0 log(s(s(X))) -> s(log(s(quot(X,s(s(0)))))) min(s(X),s(Y)) -> min(X,Y) min(X,0) -> X quot(0,s(Y)) -> 0 quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: MIN(s(X),s(Y)) -> MIN(X,Y) ->->-> Rules: log(s(0)) -> 0 log(s(s(X))) -> s(log(s(quot(X,s(s(0)))))) min(s(X),s(Y)) -> min(X,Y) min(X,0) -> X quot(0,s(Y)) -> 0 quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y))) ->->Cycle: ->->-> Pairs: QUOT(s(X),s(Y)) -> QUOT(min(X,Y),s(Y)) ->->-> Rules: log(s(0)) -> 0 log(s(s(X))) -> s(log(s(quot(X,s(s(0)))))) min(s(X),s(Y)) -> min(X,Y) min(X,0) -> X quot(0,s(Y)) -> 0 quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y))) ->->Cycle: ->->-> Pairs: LOG(s(s(X))) -> LOG(s(quot(X,s(s(0))))) ->->-> Rules: log(s(0)) -> 0 log(s(s(X))) -> s(log(s(quot(X,s(s(0)))))) min(s(X),s(Y)) -> min(X,Y) min(X,0) -> X quot(0,s(Y)) -> 0 quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y))) The problem is decomposed in 3 subproblems.
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