/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

Problem 1: 

(VAR v_NonEmpty:S X:S Y:S)
(RULES
div(0,s(Y:S)) -> 0
div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
geq(0,s(Y:S)) -> ffalse
geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
geq(X:S,0) -> ttrue
if(ffalse,X:S,Y:S) -> Y:S
if(ttrue,X:S,Y:S) -> X:S
minus(0,Y:S) -> 0
minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 DIV(s(X:S),s(Y:S)) -> DIV(minus(X:S,Y:S),s(Y:S))
 DIV(s(X:S),s(Y:S)) -> GEQ(X:S,Y:S)
 DIV(s(X:S),s(Y:S)) -> IF(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 DIV(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
 GEQ(s(X:S),s(Y:S)) -> GEQ(X:S,Y:S)
 MINUS(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)

Problem 1: 

SCC Processor:
-> Pairs:
 DIV(s(X:S),s(Y:S)) -> DIV(minus(X:S,Y:S),s(Y:S))
 DIV(s(X:S),s(Y:S)) -> GEQ(X:S,Y:S)
 DIV(s(X:S),s(Y:S)) -> IF(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 DIV(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
 GEQ(s(X:S),s(Y:S)) -> GEQ(X:S,Y:S)
 MINUS(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 MINUS(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
->->-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->->Cycle:
->->-> Pairs:
 GEQ(s(X:S),s(Y:S)) -> GEQ(X:S,Y:S)
->->-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->->Cycle:
->->-> Pairs:
 DIV(s(X:S),s(Y:S)) -> DIV(minus(X:S,Y:S),s(Y:S))
->->-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)


The problem is decomposed in 3 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 MINUS(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->Projection:
 pi(MINUS) = 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

Subterm Processor:
-> Pairs:
 GEQ(s(X:S),s(Y:S)) -> GEQ(X:S,Y:S)
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->Projection:
 pi(GEQ) = 1

Problem 1.2: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.3: 

Reduction Pairs Processor:
-> Pairs:
 DIV(s(X:S),s(Y:S)) -> DIV(minus(X:S,Y:S),s(Y:S))
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
-> Usable rules:
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[div](X1,X2) = 0
[geq](X1,X2) = 0
[if](X1,X2,X3) = 0
[minus](X1,X2) = 1
[0] = 0
[fSNonEmpty] = 0
[false] = 0
[s](X) = 2
[true] = 0
[DIV](X1,X2) = 2.X1
[GEQ](X1,X2) = 0
[IF](X1,X2,X3) = 0
[MINUS](X1,X2) = 0

Problem 1.3: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 div(0,s(Y:S)) -> 0
 div(s(X:S),s(Y:S)) -> if(geq(X:S,Y:S),s(div(minus(X:S,Y:S),s(Y:S))),0)
 geq(0,s(Y:S)) -> ffalse
 geq(s(X:S),s(Y:S)) -> geq(X:S,Y:S)
 geq(X:S,0) -> ttrue
 if(ffalse,X:S,Y:S) -> Y:S
 if(ttrue,X:S,Y:S) -> X:S
 minus(0,Y:S) -> 0
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.