details

property	value
status	complete
benchmark	17.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n023.star.cs.uiowa.edu
space	Beerendonk_07

run statistics

property	value
solver	muterm 5.18
configuration	default
runtime (wallclock)	142.754027843 seconds
cpu usage	141.11446637
max memory	6.0465152E7

stage attributes

key	value
output-size	35831
starexec-result	YES

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 add(0,x) -> x add(s(x),y) -> s(add(x,y)) cond1(true,x,y) -> cond2(gr(x,y),x,y) cond2(false,x,y) -> cond3(eq(x,y),x,y) cond2(true,x,y) -> cond1(gr(add(x,y),0),p(x),y) cond3(false,x,y) -> cond1(gr(add(x,y),0),x,p(y)) cond3(true,x,y) -> cond1(gr(add(x,y),0),p(x),y) eq(0,0) -> true eq(0,s(x)) -> false eq(s(x),0) -> false eq(s(x),s(y)) -> eq(x,y) gr(0,x) -> false gr(s(x),0) -> true gr(s(x),s(y)) -> gr(x,y) p(0) -> 0 p(s(x)) -> x ) Problem 1: Innermost Equivalent Processor: -> Rules: add(0,x) -> x add(s(x),y) -> s(add(x,y)) cond1(true,x,y) -> cond2(gr(x,y),x,y) cond2(false,x,y) -> cond3(eq(x,y),x,y) cond2(true,x,y) -> cond1(gr(add(x,y),0),p(x),y) cond3(false,x,y) -> cond1(gr(add(x,y),0),x,p(y)) cond3(true,x,y) -> cond1(gr(add(x,y),0),p(x),y) eq(0,0) -> true eq(0,s(x)) -> false eq(s(x),0) -> false eq(s(x),s(y)) -> eq(x,y) gr(0,x) -> false gr(s(x),0) -> true gr(s(x),s(y)) -> gr(x,y) 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: ADD(s(x),y) -> ADD(x,y) COND1(true,x,y) -> COND2(gr(x,y),x,y) COND1(true,x,y) -> GR(x,y) COND2(false,x,y) -> COND3(eq(x,y),x,y) COND2(false,x,y) -> EQ(x,y) COND2(true,x,y) -> ADD(x,y) COND2(true,x,y) -> COND1(gr(add(x,y),0),p(x),y) COND2(true,x,y) -> GR(add(x,y),0) COND2(true,x,y) -> P(x) COND3(false,x,y) -> ADD(x,y) COND3(false,x,y) -> COND1(gr(add(x,y),0),x,p(y)) COND3(false,x,y) -> GR(add(x,y),0) COND3(false,x,y) -> P(y) COND3(true,x,y) -> ADD(x,y) COND3(true,x,y) -> COND1(gr(add(x,y),0),p(x),y) COND3(true,x,y) -> GR(add(x,y),0) COND3(true,x,y) -> P(x) EQ(s(x),s(y)) -> EQ(x,y) GR(s(x),s(y)) -> GR(x,y) -> Rules: add(0,x) -> x add(s(x),y) -> s(add(x,y)) cond1(true,x,y) -> cond2(gr(x,y),x,y) cond2(false,x,y) -> cond3(eq(x,y),x,y) cond2(true,x,y) -> cond1(gr(add(x,y),0),p(x),y) cond3(false,x,y) -> cond1(gr(add(x,y),0),x,p(y)) cond3(true,x,y) -> cond1(gr(add(x,y),0),p(x),y) eq(0,0) -> true eq(0,s(x)) -> false eq(s(x),0) -> false eq(s(x),s(y)) -> eq(x,y) gr(0,x) -> false gr(s(x),0) -> true gr(s(x),s(y)) -> gr(x,y) p(0) -> 0 p(s(x)) -> x Problem 1: SCC Processor: -> Pairs: ADD(s(x),y) -> ADD(x,y) COND1(true,x,y) -> COND2(gr(x,y),x,y) popout

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