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TRS Stand 20472 pair #381710283
details
property
value
status
complete
benchmark
gm.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n093.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
muterm 5.18
configuration
default
runtime (wallclock)
0.0429518222809 seconds
cpu usage
0.030813812
max memory
3584000.0
stage attributes
key
value
output-size
3168
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 div(0,s(Y)) -> 0 div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y))) minus(s(X),s(Y)) -> p(minus(X,Y)) minus(X,0) -> X p(s(X)) -> X ) Problem 1: Innermost Equivalent Processor: -> Rules: div(0,s(Y)) -> 0 div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y))) minus(s(X),s(Y)) -> p(minus(X,Y)) minus(X,0) -> X p(s(X)) -> X -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: DIV(s(X),s(Y)) -> DIV(minus(X,Y),s(Y)) DIV(s(X),s(Y)) -> MINUS(X,Y) MINUS(s(X),s(Y)) -> MINUS(X,Y) MINUS(s(X),s(Y)) -> P(minus(X,Y)) -> Rules: div(0,s(Y)) -> 0 div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y))) minus(s(X),s(Y)) -> p(minus(X,Y)) minus(X,0) -> X p(s(X)) -> X Problem 1: SCC Processor: -> Pairs: DIV(s(X),s(Y)) -> DIV(minus(X,Y),s(Y)) DIV(s(X),s(Y)) -> MINUS(X,Y) MINUS(s(X),s(Y)) -> MINUS(X,Y) MINUS(s(X),s(Y)) -> P(minus(X,Y)) -> Rules: div(0,s(Y)) -> 0 div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y))) minus(s(X),s(Y)) -> p(minus(X,Y)) minus(X,0) -> X p(s(X)) -> X ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: MINUS(s(X),s(Y)) -> MINUS(X,Y) ->->-> Rules: div(0,s(Y)) -> 0 div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y))) minus(s(X),s(Y)) -> p(minus(X,Y)) minus(X,0) -> X p(s(X)) -> X ->->Cycle: ->->-> Pairs: DIV(s(X),s(Y)) -> DIV(minus(X,Y),s(Y)) ->->-> Rules: div(0,s(Y)) -> 0 div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y))) minus(s(X),s(Y)) -> p(minus(X,Y)) minus(X,0) -> X p(s(X)) -> X The problem is decomposed in 2 subproblems. Problem 1.1: Subterm Processor: -> Pairs: MINUS(s(X),s(Y)) -> MINUS(X,Y) -> Rules: div(0,s(Y)) -> 0 div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y))) minus(s(X),s(Y)) -> p(minus(X,Y)) minus(X,0) -> X p(s(X)) -> X ->Projection: pi(MINUS) = 1 Problem 1.1: SCC Processor:
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