details

property	value
status	timeout (wallclock)
benchmark	18.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
space	Beerendonk_07

run statistics

property	value
solver	muterm 5.18
configuration	default
runtime (wallclock)	301.030004025 seconds
cpu usage	308.870351979
max memory	2.7201536E7

stage attributes

unavailable

output

/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR x y) (RULES cond1(true,x,y) -> cond2(gr(x,0),x,y) cond2(false,x,y) -> cond3(gr(y,0),x,y) cond2(true,x,y) -> cond1(or(gr(x,0),gr(y,0)),p(x),y) cond3(false,x,y) -> cond1(or(gr(x,0),gr(y,0)),x,y) cond3(true,x,y) -> cond1(or(gr(x,0),gr(y,0)),x,p(y)) gr(0,x) -> false gr(s(x),0) -> true gr(s(x),s(y)) -> gr(x,y) or(false,false) -> false or(true,x) -> true or(x,true) -> true p(0) -> 0 p(s(x)) -> x ) Problem 1: Innermost Equivalent Processor: -> Rules: cond1(true,x,y) -> cond2(gr(x,0),x,y) cond2(false,x,y) -> cond3(gr(y,0),x,y) cond2(true,x,y) -> cond1(or(gr(x,0),gr(y,0)),p(x),y) cond3(false,x,y) -> cond1(or(gr(x,0),gr(y,0)),x,y) cond3(true,x,y) -> cond1(or(gr(x,0),gr(y,0)),x,p(y)) gr(0,x) -> false gr(s(x),0) -> true gr(s(x),s(y)) -> gr(x,y) or(false,false) -> false or(true,x) -> true or(x,true) -> true p(0) -> 0 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: COND1(true,x,y) -> COND2(gr(x,0),x,y) COND1(true,x,y) -> GR(x,0) COND2(false,x,y) -> COND3(gr(y,0),x,y) COND2(false,x,y) -> GR(y,0) COND2(true,x,y) -> COND1(or(gr(x,0),gr(y,0)),p(x),y) COND2(true,x,y) -> GR(x,0) COND2(true,x,y) -> GR(y,0) COND2(true,x,y) -> OR(gr(x,0),gr(y,0)) COND2(true,x,y) -> P(x) COND3(false,x,y) -> COND1(or(gr(x,0),gr(y,0)),x,y) COND3(false,x,y) -> GR(x,0) COND3(false,x,y) -> GR(y,0) COND3(false,x,y) -> OR(gr(x,0),gr(y,0)) COND3(true,x,y) -> COND1(or(gr(x,0),gr(y,0)),x,p(y)) COND3(true,x,y) -> GR(x,0) COND3(true,x,y) -> GR(y,0) COND3(true,x,y) -> OR(gr(x,0),gr(y,0)) COND3(true,x,y) -> P(y) GR(s(x),s(y)) -> GR(x,y) -> Rules: cond1(true,x,y) -> cond2(gr(x,0),x,y) cond2(false,x,y) -> cond3(gr(y,0),x,y) cond2(true,x,y) -> cond1(or(gr(x,0),gr(y,0)),p(x),y) cond3(false,x,y) -> cond1(or(gr(x,0),gr(y,0)),x,y) cond3(true,x,y) -> cond1(or(gr(x,0),gr(y,0)),x,p(y)) gr(0,x) -> false gr(s(x),0) -> true gr(s(x),s(y)) -> gr(x,y) or(false,false) -> false or(true,x) -> true or(x,true) -> true p(0) -> 0 p(s(x)) -> x Problem 1: SCC Processor: -> Pairs: COND1(true,x,y) -> COND2(gr(x,0),x,y) COND1(true,x,y) -> GR(x,0) COND2(false,x,y) -> COND3(gr(y,0),x,y) COND2(false,x,y) -> GR(y,0) COND2(true,x,y) -> COND1(or(gr(x,0),gr(y,0)),p(x),y) COND2(true,x,y) -> GR(x,0) COND2(true,x,y) -> GR(y,0) COND2(true,x,y) -> OR(gr(x,0),gr(y,0)) COND2(true,x,y) -> P(x) COND3(false,x,y) -> COND1(or(gr(x,0),gr(y,0)),x,y) COND3(false,x,y) -> GR(x,0) popout

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job log

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actions

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