/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
and(ffalse,x:S) -> ffalse
and(ttrue,ttrue) -> ttrue
and(x:S,ffalse) -> ffalse
cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
gr(0,0) -> ffalse
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:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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:
 COND(ttrue,x:S,y:S) -> AND(gr(x:S,0),gr(y:S,0))
 COND(ttrue,x:S,y:S) -> COND(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 COND(ttrue,x:S,y:S) -> GR(x:S,0)
 COND(ttrue,x:S,y:S) -> GR(y:S,0)
 COND(ttrue,x:S,y:S) -> P(x:S)
 COND(ttrue,x:S,y:S) -> P(y:S)
 GR(s(x:S),s(y:S)) -> GR(x:S,y:S)
-> Rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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:
 COND(ttrue,x:S,y:S) -> AND(gr(x:S,0),gr(y:S,0))
 COND(ttrue,x:S,y:S) -> COND(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 COND(ttrue,x:S,y:S) -> GR(x:S,0)
 COND(ttrue,x:S,y:S) -> GR(y:S,0)
 COND(ttrue,x:S,y:S) -> P(x:S)
 COND(ttrue,x:S,y:S) -> P(y:S)
 GR(s(x:S),s(y:S)) -> GR(x:S,y:S)
-> Rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 GR(s(x:S),s(y:S)) -> GR(x:S,y:S)
->->-> Rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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
->->Cycle:
->->-> Pairs:
 COND(ttrue,x:S,y:S) -> COND(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
->->-> Rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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 problem is decomposed in 2 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 GR(s(x:S),s(y:S)) -> GR(x:S,y:S)
-> Rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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
->Projection:
 pi(GR) = 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

Reduction Pairs Processor:
-> Pairs:
 COND(ttrue,x:S,y:S) -> COND(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
-> Rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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
-> Usable rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 gr(0,0) -> ffalse
 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
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[and](X1,X2) = X1
[cond](X1,X2,X3) = 0
[gr](X1,X2) = X1
[p](X) = 1/2.X
[0] = 0
[fSNonEmpty] = 0
[false] = 0
[s](X) = 2.X + 1/2
[true] = 1/2
[AND](X1,X2) = 0
[COND](X1,X2,X3) = 1/2.X1 + X2
[GR](X1,X2) = 0
[P](X) = 0

Problem 1.2: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 and(ffalse,x:S) -> ffalse
 and(ttrue,ttrue) -> ttrue
 and(x:S,ffalse) -> ffalse
 cond(ttrue,x:S,y:S) -> cond(and(gr(x:S,0),gr(y:S,0)),p(x:S),p(y:S))
 gr(0,0) -> ffalse
 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
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.