/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 q:S r:S x:S y:S)
(RULES
less(0,s(x:S)) -> ttrue
less(s(x:S),s(y:S)) -> less(x:S,y:S)
less(x:S,0) -> ffalse
minus(0,s(y:S)) -> 0
minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
minus(x:S,0) -> x:S
quotrem(0,s(y:S)) -> pair(0,0)
quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
)

Problem 1: 
Valid CTRS Processor:
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
-> The system is a deterministic 3-CTRS.

Problem 1: 

Dependency Pairs Processor:

Conditional Termination Problem 1:
-> Pairs:
 LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S)
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
-> QPairs:
 Empty
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)

Conditional Termination Problem 2:
-> Pairs:
 QUOTREM(s(x:S),s(y:S)) -> LESS(x:S,y:S)
 QUOTREM(s(x:S),s(y:S)) -> MINUS(x:S,y:S) | less(x:S,y:S) ->* ffalse
 QUOTREM(s(x:S),s(y:S)) -> QUOTREM(minus(x:S,y:S),s(y:S)) | less(x:S,y:S) ->* ffalse
-> QPairs:
 LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S)
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)


The problem is decomposed in 2 subproblems.

Problem 1.1: 

SCC Processor:
-> Pairs:
 LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S)
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
-> QPairs:
 Empty
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
-> QPairs:
 Empty
->->-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
->->Cycle:
->->-> Pairs:
 LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S)
-> QPairs:
 Empty
->->-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)


The problem is decomposed in 2 subproblems.

Problem 1.1.1: 

Conditional Subterm Processor:
-> Pairs:
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
-> QPairs:
 Empty
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
->Projection:
 pi(MINUS) = 1

Problem 1.1.1: 

SCC Processor:
-> Pairs:
 Empty
-> QPairs:
 Empty
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.1.2: 

Conditional Subterm Processor:
-> Pairs:
 LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S)
-> QPairs:
 Empty
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
->Projection:
 pi(LESS) = 1

Problem 1.1.2: 

SCC Processor:
-> Pairs:
 Empty
-> QPairs:
 Empty
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

SCC Processor:
-> Pairs:
 QUOTREM(s(x:S),s(y:S)) -> LESS(x:S,y:S)
 QUOTREM(s(x:S),s(y:S)) -> MINUS(x:S,y:S) | less(x:S,y:S) ->* ffalse
 QUOTREM(s(x:S),s(y:S)) -> QUOTREM(minus(x:S,y:S),s(y:S)) | less(x:S,y:S) ->* ffalse
-> QPairs:
 LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S)
 MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S)
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 QUOTREM(s(x:S),s(y:S)) -> QUOTREM(minus(x:S,y:S),s(y:S)) | less(x:S,y:S) ->* ffalse
-> QPairs:
 Empty
->->-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)

Problem 1.2: 

Reduction Triple Processor:
-> Pairs:
 QUOTREM(s(x:S),s(y:S)) -> QUOTREM(minus(x:S,y:S),s(y:S)) | less(x:S,y:S) ->* ffalse
-> QPairs:
 Empty
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
-> Usable rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 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:
 
[less](X1,X2) = 2.X1 + X2 + 2
[minus](X1,X2) = 2.X1
[quotrem](X1,X2) = 0
[0] = 0
[fSNonEmpty] = 0
[false] = 1
[pair](X1,X2) = 0
[s](X) = 2.X + 2
[true] = 0
[LESS](X1,X2) = 0
[MINUS](X1,X2) = 0
[QUOTREM](X1,X2) = X1

Problem 1.2: 

SCC Processor:
-> Pairs:
 Empty
-> QPairs:
 Empty
-> Rules:
 less(0,s(x:S)) -> ttrue
 less(s(x:S),s(y:S)) -> less(x:S,y:S)
 less(x:S,0) -> ffalse
 minus(0,s(y:S)) -> 0
 minus(s(x:S),s(y:S)) -> minus(x:S,y:S)
 minus(x:S,0) -> x:S
 quotrem(0,s(y:S)) -> pair(0,0)
 quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue
 quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.