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TRS Standard pair #516964651
details
property
value
status
complete
benchmark
division.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.111158132553 seconds
cpu usage
0.04614168
max memory
3878912.0
stage attributes
key
value
output-size
7115
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 ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ) Problem 1: Innermost Equivalent Processor: -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y: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: IFMINUS(ffalse,s(X:S),Y:S) -> MINUS(X:S,Y:S) LE(s(X:S),s(Y:S)) -> LE(X:S,Y:S) MINUS(s(X:S),Y:S) -> IFMINUS(le(s(X:S),Y:S),s(X:S),Y:S) MINUS(s(X:S),Y:S) -> LE(s(X:S),Y:S) QUOT(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S) QUOT(s(X:S),s(Y:S)) -> QUOT(minus(X:S,Y:S),s(Y:S)) -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) Problem 1: SCC Processor: -> Pairs: IFMINUS(ffalse,s(X:S),Y:S) -> MINUS(X:S,Y:S) LE(s(X:S),s(Y:S)) -> LE(X:S,Y:S) MINUS(s(X:S),Y:S) -> IFMINUS(le(s(X:S),Y:S),s(X:S),Y:S) MINUS(s(X:S),Y:S) -> LE(s(X:S),Y:S) QUOT(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S) QUOT(s(X:S),s(Y:S)) -> QUOT(minus(X:S,Y:S),s(Y:S)) -> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: LE(s(X:S),s(Y:S)) -> LE(X:S,Y:S) ->->-> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S)) ifMinus(ttrue,s(X:S),Y:S) -> 0 le(0,Y:S) -> ttrue le(s(X:S),0) -> ffalse le(s(X:S),s(Y:S)) -> le(X:S,Y:S) minus(0,Y:S) -> 0 minus(s(X:S),Y:S) -> ifMinus(le(s(X:S),Y:S),s(X:S),Y:S) quot(0,s(Y:S)) -> 0 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S))) ->->Cycle: ->->-> Pairs: IFMINUS(ffalse,s(X:S),Y:S) -> MINUS(X:S,Y:S) MINUS(s(X:S),Y:S) -> IFMINUS(le(s(X:S),Y:S),s(X:S),Y:S) ->->-> Rules: ifMinus(ffalse,s(X:S),Y:S) -> s(minus(X:S,Y:S))
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