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TRS Standard pair #487072639
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
n178.star.cs.uiowa.edu
space
Beerendonk_07
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
289.265 seconds
cpu usage
287.914
user time
272.885
system time
15.0299
max virtual memory
761792.0
max residence set size
48736.0
stage attributes
key
value
starexec-result
YES
output
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) 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)
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