/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 v_NonEmpty:S x:S y:S z:S)
(RULES
=(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v))
=(.(x:S,y:S),nil) -> ffalse
=(nil,.(y:S,z:S)) -> ffalse
=(nil,nil) -> ttrue
del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S)
f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S)))
f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S))
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v))
 =(.(x:S,y:S),nil) -> ffalse
 =(nil,.(y:S,z:S)) -> ffalse
 =(nil,nil) -> ttrue
 del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S)
 f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S)))
 f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S))
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 DEL(.(x:S,.(y:S,z:S))) -> =#(x:S,y:S)
 DEL(.(x:S,.(y:S,z:S))) -> F(=(x:S,y:S),x:S,y:S,z:S)
 F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
 F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
-> Rules:
 =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v))
 =(.(x:S,y:S),nil) -> ffalse
 =(nil,.(y:S,z:S)) -> ffalse
 =(nil,nil) -> ttrue
 del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S)
 f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S)))
 f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S))

Problem 1: 

SCC Processor:
-> Pairs:
 DEL(.(x:S,.(y:S,z:S))) -> =#(x:S,y:S)
 DEL(.(x:S,.(y:S,z:S))) -> F(=(x:S,y:S),x:S,y:S,z:S)
 F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
 F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
-> Rules:
 =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v))
 =(.(x:S,y:S),nil) -> ffalse
 =(nil,.(y:S,z:S)) -> ffalse
 =(nil,nil) -> ttrue
 del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S)
 f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S)))
 f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 DEL(.(x:S,.(y:S,z:S))) -> F(=(x:S,y:S),x:S,y:S,z:S)
 F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
 F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
->->-> Rules:
 =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v))
 =(.(x:S,y:S),nil) -> ffalse
 =(nil,.(y:S,z:S)) -> ffalse
 =(nil,nil) -> ttrue
 del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S)
 f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S)))
 f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S))

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 DEL(.(x:S,.(y:S,z:S))) -> F(=(x:S,y:S),x:S,y:S,z:S)
 F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
 F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
-> Rules:
 =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v))
 =(.(x:S,y:S),nil) -> ffalse
 =(nil,.(y:S,z:S)) -> ffalse
 =(nil,nil) -> ttrue
 del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S)
 f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S)))
 f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S))
-> Usable rules:
 =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v))
 =(.(x:S,y:S),nil) -> ffalse
 =(nil,.(y:S,z:S)) -> ffalse
 =(nil,nil) -> ttrue
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[=](X1,X2) = 2.X1 + 2.X2 + 2
[del](X) = 0
[f](X1,X2,X3,X4) = 0
[.](X1,X2) = 2.X1 + 2.X2 + 2
[and](X1,X2) = X2 + 2
[fSNonEmpty] = 0
[false] = 0
[nil] = 2
[true] = 2
[u] = 2
[v] = 1
[=#](X1,X2) = 0
[DEL](X) = X
[F](X1,X2,X3,X4) = X2 + 2.X3 + 2.X4 + 2

Problem 1: 

SCC Processor:
-> Pairs:
 F(ffalse,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
 F(ttrue,x:S,y:S,z:S) -> DEL(.(y:S,z:S))
-> Rules:
 =(.(x:S,y:S),.(u,v)) -> and(=(x:S,u),=(y:S,v))
 =(.(x:S,y:S),nil) -> ffalse
 =(nil,.(y:S,z:S)) -> ffalse
 =(nil,nil) -> ttrue
 del(.(x:S,.(y:S,z:S))) -> f(=(x:S,y:S),x:S,y:S,z:S)
 f(ffalse,x:S,y:S,z:S) -> .(x:S,del(.(y:S,z:S)))
 f(ttrue,x:S,y:S,z:S) -> del(.(y:S,z:S))
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.