/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 q r x y)
(RULES
less(0,s(x)) -> true
less(s(x),s(y)) -> less(x,y)
less(x,0) -> false
minus(0,s(y)) -> 0
minus(s(x),s(y)) -> minus(x,y)
minus(x,0) -> x
quotrem(0,s(y)) -> pair(0,0)
quotrem(s(x),s(y)) -> pair(0,s(x)) | less(x,y) -> true
quotrem(s(x),s(y)) -> pair(s(q),r) | less(x,y) -> false, quotrem(minus(x,y),s(y)) -> pair(q,r)
)

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

Problem 1: 

Dependency Pairs Processor:

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

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


The problem is decomposed in 2 subproblems.

Problem 1.1: 

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


The problem is decomposed in 2 subproblems.

Problem 1.1.1: 

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

Problem 1.1.1: 

SCC Processor:
-> Pairs:
 Empty
-> QPairs:
 Empty
-> Rules:
 less(0,s(x)) -> true
 less(s(x),s(y)) -> less(x,y)
 less(x,0) -> false
 minus(0,s(y)) -> 0
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 quotrem(0,s(y)) -> pair(0,0)
 quotrem(s(x),s(y)) -> pair(0,s(x)) | less(x,y) -> true
 quotrem(s(x),s(y)) -> pair(s(q),r) | less(x,y) -> false, quotrem(minus(x,y),s(y)) -> pair(q,r)
->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(y)) -> LESS(x,y)
-> QPairs:
 Empty
-> Rules:
 less(0,s(x)) -> true
 less(s(x),s(y)) -> less(x,y)
 less(x,0) -> false
 minus(0,s(y)) -> 0
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 quotrem(0,s(y)) -> pair(0,0)
 quotrem(s(x),s(y)) -> pair(0,s(x)) | less(x,y) -> true
 quotrem(s(x),s(y)) -> pair(s(q),r) | less(x,y) -> false, quotrem(minus(x,y),s(y)) -> pair(q,r)
->Projection:
 pi(LESS) = 1

Problem 1.1.2: 

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

The problem is finite.

Problem 1.2: 

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

Problem 1.2: 

Reduction Triple Processor:
-> Pairs:
 QUOTREM(s(x),s(y)) -> QUOTREM(minus(x,y),s(y)) | less(x,y) -> false
-> QPairs:
 Empty
-> Rules:
 less(0,s(x)) -> true
 less(s(x),s(y)) -> less(x,y)
 less(x,0) -> false
 minus(0,s(y)) -> 0
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 quotrem(0,s(y)) -> pair(0,0)
 quotrem(s(x),s(y)) -> pair(0,s(x)) | less(x,y) -> true
 quotrem(s(x),s(y)) -> pair(s(q),r) | less(x,y) -> false, quotrem(minus(x,y),s(y)) -> pair(q,r)
-> Usable rules:
 less(0,s(x)) -> true
 less(s(x),s(y)) -> less(x,y)
 less(x,0) -> false
 minus(0,s(y)) -> 0
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
->Interpretation type:
 Linear
->Coefficients:
 Integer Numbers
->Dimension:
 1
->Convex Domain:
 D[(x_1)] = ((1+x_1 >= 0) /\ (-1+x_1 >= 0))
->Interpretation:
 
[delta] = 1
[less](x_1,x_2) = 1
[minus](x_1,x_2) = x_1
[0] = 1
[false] = 1
[s](x_1) = 1+x_1
[true] = 1
[QUOTREM](x_1,x_2) = x_1

Problem 1.2: 

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

The problem is finite.