/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)
(RULES
cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
div2(0) -> 0
div2(s(0)) -> 0
div2(s(s(x:S))) -> s(div2(x:S))
even(0) -> ttrue
even(s(0)) -> ffalse
even(s(s(x:S))) -> even(x:S)
neq(0,0) -> ffalse
neq(0,s(x:S)) -> ttrue
neq(s(x:S),0) -> ttrue
neq(s(x:S),s(y)) -> neq(x:S,y)
p(0) -> 0
p(s(x:S)) -> x:S
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 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:
 COND1(ttrue,x:S) -> COND2(even(x:S),x:S)
 COND1(ttrue,x:S) -> EVEN(x:S)
 COND2(ffalse,x:S) -> COND1(neq(x:S,0),p(x:S))
 COND2(ffalse,x:S) -> NEQ(x:S,0)
 COND2(ffalse,x:S) -> P(x:S)
 COND2(ttrue,x:S) -> COND1(neq(x:S,0),div2(x:S))
 COND2(ttrue,x:S) -> DIV2(x:S)
 COND2(ttrue,x:S) -> NEQ(x:S,0)
 DIV2(s(s(x:S))) -> DIV2(x:S)
 EVEN(s(s(x:S))) -> EVEN(x:S)
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S

Problem 1: 

SCC Processor:
-> Pairs:
 COND1(ttrue,x:S) -> COND2(even(x:S),x:S)
 COND1(ttrue,x:S) -> EVEN(x:S)
 COND2(ffalse,x:S) -> COND1(neq(x:S,0),p(x:S))
 COND2(ffalse,x:S) -> NEQ(x:S,0)
 COND2(ffalse,x:S) -> P(x:S)
 COND2(ttrue,x:S) -> COND1(neq(x:S,0),div2(x:S))
 COND2(ttrue,x:S) -> DIV2(x:S)
 COND2(ttrue,x:S) -> NEQ(x:S,0)
 DIV2(s(s(x:S))) -> DIV2(x:S)
 EVEN(s(s(x:S))) -> EVEN(x:S)
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 EVEN(s(s(x:S))) -> EVEN(x:S)
->->-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
->->Cycle:
->->-> Pairs:
 DIV2(s(s(x:S))) -> DIV2(x:S)
->->-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
->->Cycle:
->->-> Pairs:
 COND1(ttrue,x:S) -> COND2(even(x:S),x:S)
 COND2(ffalse,x:S) -> COND1(neq(x:S,0),p(x:S))
 COND2(ttrue,x:S) -> COND1(neq(x:S,0),div2(x:S))
->->-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S


The problem is decomposed in 3 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 EVEN(s(s(x:S))) -> EVEN(x:S)
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Projection:
 pi(EVEN) = 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 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: 

Subterm Processor:
-> Pairs:
 DIV2(s(s(x:S))) -> DIV2(x:S)
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Projection:
 pi(DIV2) = 1

Problem 1.2: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.3: 

Reduction Pairs Processor:
-> Pairs:
 COND1(ttrue,x:S) -> COND2(even(x:S),x:S)
 COND2(ffalse,x:S) -> COND1(neq(x:S,0),p(x:S))
 COND2(ttrue,x:S) -> COND1(neq(x:S,0),div2(x:S))
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
-> Usable rules:
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[cond1](X1,X2) = 0
[cond2](X1,X2) = 0
[div2](X) = 1/2.X
[even](X) = 1/2
[neq](X1,X2) = 1/2.X1 + X2
[p](X) = 1/2.X
[0] = 0
[fSNonEmpty] = 0
[false] = 0
[s](X) = 2.X + 2
[true] = 1/2
[y] = 1/2
[COND1](X1,X2) = 2.X1 + 2.X2
[COND2](X1,X2) = X1 + 2.X2
[DIV2](X) = 0
[EVEN](X) = 0
[NEQ](X1,X2) = 0
[P](X) = 0

Problem 1.3: 

SCC Processor:
-> Pairs:
 COND2(ffalse,x:S) -> COND1(neq(x:S,0),p(x:S))
 COND2(ttrue,x:S) -> COND1(neq(x:S,0),div2(x:S))
-> Rules:
 cond1(ttrue,x:S) -> cond2(even(x:S),x:S)
 cond2(ffalse,x:S) -> cond1(neq(x:S,0),p(x:S))
 cond2(ttrue,x:S) -> cond1(neq(x:S,0),div2(x:S))
 div2(0) -> 0
 div2(s(0)) -> 0
 div2(s(s(x:S))) -> s(div2(x:S))
 even(0) -> ttrue
 even(s(0)) -> ffalse
 even(s(s(x:S))) -> even(x:S)
 neq(0,0) -> ffalse
 neq(0,s(x:S)) -> ttrue
 neq(s(x:S),0) -> ttrue
 neq(s(x:S),s(y)) -> neq(x:S,y)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.