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TRS Standard pair #516967161
details
property
value
status
complete
benchmark
#3.5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n060.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.118302106857 seconds
cpu usage
0.04712009
max memory
3874816.0
stage attributes
key
value
output-size
7285
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 v_NonEmpty:S x:S y:S) (RULES if_mod(ffalse,s(x:S),s(y:S)) -> s(x:S) if_mod(ttrue,s(x:S),s(y:S)) -> mod(minus(x:S,y:S),s(y:S)) le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S mod(0,y:S) -> 0 mod(s(x:S),0) -> 0 mod(s(x:S),s(y:S)) -> if_mod(le(y:S,x:S),s(x:S),s(y:S)) ) Problem 1: Innermost Equivalent Processor: -> Rules: if_mod(ffalse,s(x:S),s(y:S)) -> s(x:S) if_mod(ttrue,s(x:S),s(y:S)) -> mod(minus(x:S,y:S),s(y:S)) le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S mod(0,y:S) -> 0 mod(s(x:S),0) -> 0 mod(s(x:S),s(y:S)) -> if_mod(le(y:S,x:S),s(x:S),s(y:S)) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: IF_MOD(ttrue,s(x:S),s(y:S)) -> MINUS(x:S,y:S) IF_MOD(ttrue,s(x:S),s(y:S)) -> MOD(minus(x:S,y:S),s(y:S)) LE(s(x:S),s(y:S)) -> LE(x:S,y:S) MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S) MOD(s(x:S),s(y:S)) -> IF_MOD(le(y:S,x:S),s(x:S),s(y:S)) MOD(s(x:S),s(y:S)) -> LE(y:S,x:S) -> Rules: if_mod(ffalse,s(x:S),s(y:S)) -> s(x:S) if_mod(ttrue,s(x:S),s(y:S)) -> mod(minus(x:S,y:S),s(y:S)) le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S mod(0,y:S) -> 0 mod(s(x:S),0) -> 0 mod(s(x:S),s(y:S)) -> if_mod(le(y:S,x:S),s(x:S),s(y:S)) Problem 1: SCC Processor: -> Pairs: IF_MOD(ttrue,s(x:S),s(y:S)) -> MINUS(x:S,y:S) IF_MOD(ttrue,s(x:S),s(y:S)) -> MOD(minus(x:S,y:S),s(y:S)) LE(s(x:S),s(y:S)) -> LE(x:S,y:S) MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S) MOD(s(x:S),s(y:S)) -> IF_MOD(le(y:S,x:S),s(x:S),s(y:S)) MOD(s(x:S),s(y:S)) -> LE(y:S,x:S) -> Rules: if_mod(ffalse,s(x:S),s(y:S)) -> s(x:S) if_mod(ttrue,s(x:S),s(y:S)) -> mod(minus(x:S,y:S),s(y:S)) le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S mod(0,y:S) -> 0 mod(s(x:S),0) -> 0 mod(s(x:S),s(y:S)) -> if_mod(le(y:S,x:S),s(x:S),s(y:S)) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S) ->->-> Rules: if_mod(ffalse,s(x:S),s(y:S)) -> s(x:S) if_mod(ttrue,s(x:S),s(y:S)) -> mod(minus(x:S,y:S),s(y:S)) le(0,y:S) -> ttrue le(s(x:S),0) -> ffalse le(s(x:S),s(y:S)) -> le(x:S,y:S) minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S mod(0,y:S) -> 0 mod(s(x:S),0) -> 0 mod(s(x:S),s(y:S)) -> if_mod(le(y:S,x:S),s(x:S),s(y:S)) ->->Cycle:
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