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	n046.star.cs.uiowa.edu
space	Beerendonk_07

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	292.174968004 seconds
cpu usage	308.834511463
max memory	1.11112192E8

stage attributes

key	value
output-size	43487
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 add(0,x:S) -> x:S add(s(x:S),y:S) -> s(add(x:S,y:S)) cond1(ttrue,x:S,y:S) -> cond2(gr(x:S,y:S),x:S,y:S) cond2(ffalse,x:S,y:S) -> cond3(eq(x:S,y:S),x:S,y:S) cond2(ttrue,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),p(x:S),y:S) cond3(ffalse,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),x:S,p(y:S)) cond3(ttrue,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),p(x:S),y:S) eq(0,0) -> ttrue eq(0,s(x:S)) -> ffalse eq(s(x:S),0) -> ffalse eq(s(x:S),s(y:S)) -> eq(x:S,y:S) gr(0,x:S) -> ffalse gr(s(x:S),0) -> ttrue gr(s(x:S),s(y:S)) -> gr(x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S ) Problem 1: Innermost Equivalent Processor: -> Rules: add(0,x:S) -> x:S add(s(x:S),y:S) -> s(add(x:S,y:S)) cond1(ttrue,x:S,y:S) -> cond2(gr(x:S,y:S),x:S,y:S) cond2(ffalse,x:S,y:S) -> cond3(eq(x:S,y:S),x:S,y:S) cond2(ttrue,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),p(x:S),y:S) cond3(ffalse,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),x:S,p(y:S)) cond3(ttrue,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),p(x:S),y:S) eq(0,0) -> ttrue eq(0,s(x:S)) -> ffalse eq(s(x:S),0) -> ffalse eq(s(x:S),s(y:S)) -> eq(x:S,y:S) gr(0,x:S) -> ffalse gr(s(x:S),0) -> ttrue gr(s(x:S),s(y:S)) -> gr(x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S -> 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:S),y:S) -> ADD(x:S,y:S) COND1(ttrue,x:S,y:S) -> COND2(gr(x:S,y:S),x:S,y:S) COND1(ttrue,x:S,y:S) -> GR(x:S,y:S) COND2(ffalse,x:S,y:S) -> COND3(eq(x:S,y:S),x:S,y:S) COND2(ffalse,x:S,y:S) -> EQ(x:S,y:S) COND2(ttrue,x:S,y:S) -> ADD(x:S,y:S) COND2(ttrue,x:S,y:S) -> COND1(gr(add(x:S,y:S),0),p(x:S),y:S) COND2(ttrue,x:S,y:S) -> GR(add(x:S,y:S),0) COND2(ttrue,x:S,y:S) -> P(x:S) COND3(ffalse,x:S,y:S) -> ADD(x:S,y:S) COND3(ffalse,x:S,y:S) -> COND1(gr(add(x:S,y:S),0),x:S,p(y:S)) COND3(ffalse,x:S,y:S) -> GR(add(x:S,y:S),0) COND3(ffalse,x:S,y:S) -> P(y:S) COND3(ttrue,x:S,y:S) -> ADD(x:S,y:S) COND3(ttrue,x:S,y:S) -> COND1(gr(add(x:S,y:S),0),p(x:S),y:S) COND3(ttrue,x:S,y:S) -> GR(add(x:S,y:S),0) COND3(ttrue,x:S,y:S) -> P(x:S) EQ(s(x:S),s(y:S)) -> EQ(x:S,y:S) GR(s(x:S),s(y:S)) -> GR(x:S,y:S) -> Rules: add(0,x:S) -> x:S add(s(x:S),y:S) -> s(add(x:S,y:S)) cond1(ttrue,x:S,y:S) -> cond2(gr(x:S,y:S),x:S,y:S) cond2(ffalse,x:S,y:S) -> cond3(eq(x:S,y:S),x:S,y:S) cond2(ttrue,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),p(x:S),y:S) cond3(ffalse,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),x:S,p(y:S)) cond3(ttrue,x:S,y:S) -> cond1(gr(add(x:S,y:S),0),p(x:S),y:S) eq(0,0) -> ttrue eq(0,s(x:S)) -> ffalse eq(s(x:S),0) -> ffalse eq(s(x:S),s(y:S)) -> eq(x:S,y:S) gr(0,x:S) -> ffalse gr(s(x:S),0) -> ttrue gr(s(x:S),s(y:S)) -> gr(x:S,y:S) p(0) -> 0 p(s(x:S)) -> x:S Problem 1: SCC Processor: -> Pairs: ADD(s(x:S),y:S) -> ADD(x:S,y:S) COND1(ttrue,x:S,y:S) -> COND2(gr(x:S,y:S),x:S,y:S) popout

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