/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
minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
minus(x:S,0) -> x:S
quot(0,s(y:S)) -> 0
quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),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:
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
 QUOT(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
 QUOT(s(x:S),s(y:S)) -> QUOT(minus(x:S,y:S),s(y:S))
-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))

Problem 1: 

SCC Processor:
-> Pairs:
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
 QUOT(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
 QUOT(s(x:S),s(y:S)) -> QUOT(minus(x:S,y:S),s(y:S))
-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
->->-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))
->->Cycle:
->->-> Pairs:
 QUOT(s(x:S),s(y:S)) -> QUOT(minus(x:S,y:S),s(y:S))
->->-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))


The problem is decomposed in 2 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))
->Projection:
 pi(MINUS) = 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

Reduction Pairs Processor:
-> Pairs:
 QUOT(s(x:S),s(y:S)) -> QUOT(minus(x:S,y:S),s(y:S))
-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))
-> Usable rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[minus](X1,X2) = 2.X1 + 1
[quot](X1,X2) = 0
[0] = 0
[fSNonEmpty] = 0
[s](X) = 2.X + 2
[MINUS](X1,X2) = 0
[QUOT](X1,X2) = 2.X1

Problem 1.2: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quot(0,s(y:S)) -> 0
 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S)))
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.