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


--------------------------------------------------------------------------------


YES

Problem 1: 

(VAR x y z)
(RULES
=(.(x,y),.(u,v)) -> and(=(x,u),=(y,v))
=(.(x,y),nil) -> false
=(nil,.(y,z)) -> false
=(nil,nil) -> true
del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
f(false,x,y,z) -> .(x,del(.(y,z)))
f(true,x,y,z) -> del(.(y,z))
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 =(.(x,y),.(u,v)) -> and(=(x,u),=(y,v))
 =(.(x,y),nil) -> false
 =(nil,.(y,z)) -> false
 =(nil,nil) -> true
 del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
 f(false,x,y,z) -> .(x,del(.(y,z)))
 f(true,x,y,z) -> del(.(y,z))
-> 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,.(y,z))) -> =#(x,y)
 DEL(.(x,.(y,z))) -> F(=(x,y),x,y,z)
 F(false,x,y,z) -> DEL(.(y,z))
 F(true,x,y,z) -> DEL(.(y,z))
-> Rules:
 =(.(x,y),.(u,v)) -> and(=(x,u),=(y,v))
 =(.(x,y),nil) -> false
 =(nil,.(y,z)) -> false
 =(nil,nil) -> true
 del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
 f(false,x,y,z) -> .(x,del(.(y,z)))
 f(true,x,y,z) -> del(.(y,z))

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.