Term Rewriting System R:
[x, y, z]
del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
f(true, x, y, z) -> del(.(y, z))
f(false, x, y, z) -> .(x, del(.(y, z)))
=(nil, nil) -> true
=(.(x, y), nil) -> false
=(nil, .(y, z)) -> false
=(.(x, y), .(u, v)) -> and(=(x, u), =(y, v))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)
DEL(.(x, .(y, z))) -> ='(x, y)
F(true, x, y, z) -> DEL(.(y, z))
F(false, x, y, z) -> DEL(.(y, z))
='(.(x, y), .(u, v)) -> ='(x, u)
='(.(x, y), .(u, v)) -> ='(y, v)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

F(false, x, y, z) -> DEL(.(y, z))
F(true, x, y, z) -> DEL(.(y, z))
DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)


Rules:


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


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

DEL(.(x, .(y, z))) -> F(=(x, y), x, y, z)
four new Dependency Pairs are created:

DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
DEL(.(.(x'', y''), .(.(u, v), z))) -> F(and(=(x'', u), =(y'', v)), .(x'', y''), .(u, v), z)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Instantiation Transformation


Dependency Pairs:

DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
F(true, x, y, z) -> DEL(.(y, z))
DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
F(false, x, y, z) -> DEL(.(y, z))


Rules:


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


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(true, x, y, z) -> DEL(.(y, z))
one new Dependency Pair is created:

F(true, nil, nil, z'') -> DEL(.(nil, z''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 3
                 ↳Instantiation Transformation


Dependency Pairs:

DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
F(true, nil, nil, z'') -> DEL(.(nil, z''))
DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
F(false, x, y, z) -> DEL(.(y, z))
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)


Rules:


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


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(false, x, y, z) -> DEL(.(y, z))
two new Dependency Pairs are created:

F(false, .(x'''', y''''), nil, z'') -> DEL(.(nil, z''))
F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 4
                 ↳Forward Instantiation Transformation


Dependency Pairs:

F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
F(true, nil, nil, z'') -> DEL(.(nil, z''))
DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
F(false, .(x'''', y''''), nil, z'') -> DEL(.(nil, z''))
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)


Rules:


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


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(true, nil, nil, z'') -> DEL(.(nil, z''))
two new Dependency Pairs are created:

F(true, nil, nil, .(nil, z''')) -> DEL(.(nil, .(nil, z''')))
F(true, nil, nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 5
                 ↳Forward Instantiation Transformation


Dependency Pairs:

DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
F(true, nil, nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
F(true, nil, nil, .(nil, z''')) -> DEL(.(nil, .(nil, z''')))
DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
F(false, .(x'''', y''''), nil, z'') -> DEL(.(nil, z''))
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))


Rules:


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


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

DEL(.(nil, .(nil, z))) -> F(true, nil, nil, z)
two new Dependency Pairs are created:

DEL(.(nil, .(nil, .(nil, z''''')))) -> F(true, nil, nil, .(nil, z'''''))
DEL(.(nil, .(nil, .(.(y'''''', z''''''), z'0'')))) -> F(true, nil, nil, .(.(y'''''', z''''''), z'0''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 6
                 ↳Forward Instantiation Transformation


Dependency Pairs:

F(true, nil, nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(nil, .(nil, .(.(y'''''', z''''''), z'0'')))) -> F(true, nil, nil, .(.(y'''''', z''''''), z'0''))
F(true, nil, nil, .(nil, z''')) -> DEL(.(nil, .(nil, z''')))
DEL(.(nil, .(nil, .(nil, z''''')))) -> F(true, nil, nil, .(nil, z'''''))
F(false, .(x'''', y''''), nil, z'') -> DEL(.(nil, z''))
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)


Rules:


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


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(false, .(x'''', y''''), nil, z'') -> DEL(.(nil, z''))
three new Dependency Pairs are created:

F(false, .(x'''', y''''), nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
F(false, .(x'''', y''''), nil, .(nil, .(nil, z'''''''))) -> DEL(.(nil, .(nil, .(nil, z'''''''))))
F(false, .(x'''', y''''), nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))) -> DEL(.(nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 7
                 ↳Forward Instantiation Transformation


Dependency Pairs:

F(false, .(x'''', y''''), nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))) -> DEL(.(nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))))
DEL(.(nil, .(nil, .(.(y'''''', z''''''), z'0'')))) -> F(true, nil, nil, .(.(y'''''', z''''''), z'0''))
F(true, nil, nil, .(nil, z''')) -> DEL(.(nil, .(nil, z''')))
DEL(.(nil, .(nil, .(nil, z''''')))) -> F(true, nil, nil, .(nil, z'''''))
F(false, .(x'''', y''''), nil, .(nil, .(nil, z'''''''))) -> DEL(.(nil, .(nil, .(nil, z'''''''))))
F(false, .(x'''', y''''), nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
F(true, nil, nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))


Rules:


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


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

DEL(.(.(x'', y''), .(nil, z))) -> F(false, .(x'', y''), nil, z)
three new Dependency Pairs are created:

DEL(.(.(x''', y'''), .(nil, .(.(y''''''', z''''''), z'0'')))) -> F(false, .(x''', y'''), nil, .(.(y''''''', z''''''), z'0''))
DEL(.(.(x''', y'''), .(nil, .(nil, .(nil, z'''''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(nil, z''''''''')))
DEL(.(.(x''', y'''), .(nil, .(nil, .(.(y'''''''''', z''''''''''), z'0''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(.(y'''''''''', z''''''''''), z'0'''''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 8
                 ↳Forward Instantiation Transformation


Dependency Pairs:

DEL(.(.(x''', y'''), .(nil, .(nil, .(.(y'''''''''', z''''''''''), z'0''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(.(y'''''''''', z''''''''''), z'0'''''')))
F(true, nil, nil, .(nil, z''')) -> DEL(.(nil, .(nil, z''')))
DEL(.(nil, .(nil, .(nil, z''''')))) -> F(true, nil, nil, .(nil, z'''''))
F(false, .(x'''', y''''), nil, .(nil, .(nil, z'''''''))) -> DEL(.(nil, .(nil, .(nil, z'''''''))))
DEL(.(.(x''', y'''), .(nil, .(nil, .(nil, z'''''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(nil, z''''''''')))
F(false, .(x'''', y''''), nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(.(x''', y'''), .(nil, .(.(y''''''', z''''''), z'0'')))) -> F(false, .(x''', y'''), nil, .(.(y''''''', z''''''), z'0''))
F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
F(true, nil, nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(nil, .(nil, .(.(y'''''', z''''''), z'0'')))) -> F(true, nil, nil, .(.(y'''''', z''''''), z'0''))
F(false, .(x'''', y''''), nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))) -> DEL(.(nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))))


Rules:


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


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(false, nil, .(y'''', z''''), z'') -> DEL(.(.(y'''', z''''), z''))
three new Dependency Pairs are created:

F(false, nil, .(y'''''', z'''''), .(nil, .(.(y''''''''', z''''''''), z'0''''))) -> DEL(.(.(y'''''', z'''''), .(nil, .(.(y''''''''', z''''''''), z'0''''))))
F(false, nil, .(y'''''', z'''''), .(nil, .(nil, .(nil, z''''''''''')))) -> DEL(.(.(y'''''', z'''''), .(nil, .(nil, .(nil, z''''''''''')))))
F(false, nil, .(y'''''', z'''''), .(nil, .(nil, .(.(y'''''''''''', z''''''''''''), z'0'''''''')))) -> DEL(.(.(y'''''', z'''''), .(nil, .(nil, .(.(y'''''''''''', z''''''''''''), z'0'''''''')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 9
                 ↳Forward Instantiation Transformation


Dependency Pairs:

F(false, nil, .(y'''''', z'''''), .(nil, .(nil, .(.(y'''''''''''', z''''''''''''), z'0'''''''')))) -> DEL(.(.(y'''''', z'''''), .(nil, .(nil, .(.(y'''''''''''', z''''''''''''), z'0'''''''')))))
F(true, nil, nil, .(nil, z''')) -> DEL(.(nil, .(nil, z''')))
DEL(.(nil, .(nil, .(nil, z''''')))) -> F(true, nil, nil, .(nil, z'''''))
F(false, .(x'''', y''''), nil, .(nil, .(nil, z'''''''))) -> DEL(.(nil, .(nil, .(nil, z'''''''))))
DEL(.(.(x''', y'''), .(nil, .(nil, .(nil, z'''''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(nil, z''''''''')))
F(false, nil, .(y'''''', z'''''), .(nil, .(nil, .(nil, z''''''''''')))) -> DEL(.(.(y'''''', z'''''), .(nil, .(nil, .(nil, z''''''''''')))))
F(false, .(x'''', y''''), nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(.(x''', y'''), .(nil, .(.(y''''''', z''''''), z'0'')))) -> F(false, .(x''', y'''), nil, .(.(y''''''', z''''''), z'0''))
F(false, nil, .(y'''''', z'''''), .(nil, .(.(y''''''''', z''''''''), z'0''''))) -> DEL(.(.(y'''''', z'''''), .(nil, .(.(y''''''''', z''''''''), z'0''''))))
DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
F(true, nil, nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(nil, .(nil, .(.(y'''''', z''''''), z'0'')))) -> F(true, nil, nil, .(.(y'''''', z''''''), z'0''))
F(false, .(x'''', y''''), nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))) -> DEL(.(nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))))
DEL(.(.(x''', y'''), .(nil, .(nil, .(.(y'''''''''', z''''''''''), z'0''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(.(y'''''''''', z''''''''''), z'0'''''')))


Rules:


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


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

DEL(.(nil, .(.(y'', z''), z))) -> F(false, nil, .(y'', z''), z)
three new Dependency Pairs are created:

DEL(.(nil, .(.(y''', z'''), .(nil, .(.(y''''''''''', z''''''''''), z'0''''''))))) -> F(false, nil, .(y''', z'''), .(nil, .(.(y''''''''''', z''''''''''), z'0'''''')))
DEL(.(nil, .(.(y''', z'''), .(nil, .(nil, .(nil, z''''''''''''')))))) -> F(false, nil, .(y''', z'''), .(nil, .(nil, .(nil, z'''''''''''''))))
DEL(.(nil, .(.(y''', z'''), .(nil, .(nil, .(.(y'''''''''''''', z''''''''''''''), z'0'''''''''')))))) -> F(false, nil, .(y''', z'''), .(nil, .(nil, .(.(y'''''''''''''', z''''''''''''''), z'0''''''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 10
                 ↳Polynomial Ordering


Dependency Pairs:

DEL(.(nil, .(.(y''', z'''), .(nil, .(nil, .(.(y'''''''''''''', z''''''''''''''), z'0'''''''''')))))) -> F(false, nil, .(y''', z'''), .(nil, .(nil, .(.(y'''''''''''''', z''''''''''''''), z'0''''''''''))))
F(true, nil, nil, .(nil, z''')) -> DEL(.(nil, .(nil, z''')))
DEL(.(nil, .(nil, .(nil, z''''')))) -> F(true, nil, nil, .(nil, z'''''))
F(false, .(x'''', y''''), nil, .(nil, .(nil, z'''''''))) -> DEL(.(nil, .(nil, .(nil, z'''''''))))
DEL(.(.(x''', y'''), .(nil, .(nil, .(nil, z'''''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(nil, z''''''''')))
F(false, nil, .(y'''''', z'''''), .(nil, .(nil, .(nil, z''''''''''')))) -> DEL(.(.(y'''''', z'''''), .(nil, .(nil, .(nil, z''''''''''')))))
DEL(.(nil, .(.(y''', z'''), .(nil, .(nil, .(nil, z''''''''''''')))))) -> F(false, nil, .(y''', z'''), .(nil, .(nil, .(nil, z'''''''''''''))))
F(false, .(x'''', y''''), nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(.(x''', y'''), .(nil, .(.(y''''''', z''''''), z'0'')))) -> F(false, .(x''', y'''), nil, .(.(y''''''', z''''''), z'0''))
F(false, nil, .(y'''''', z'''''), .(nil, .(.(y''''''''', z''''''''), z'0''''))) -> DEL(.(.(y'''''', z'''''), .(nil, .(.(y''''''''', z''''''''), z'0''''))))
DEL(.(nil, .(.(y''', z'''), .(nil, .(.(y''''''''''', z''''''''''), z'0''''''))))) -> F(false, nil, .(y''', z'''), .(nil, .(.(y''''''''''', z''''''''''), z'0'''''')))
F(true, nil, nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(nil, .(nil, .(.(y'''''', z''''''), z'0'')))) -> F(true, nil, nil, .(.(y'''''', z''''''), z'0''))
F(false, .(x'''', y''''), nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))) -> DEL(.(nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))))
DEL(.(.(x''', y'''), .(nil, .(nil, .(.(y'''''''''', z''''''''''), z'0''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(.(y'''''''''', z''''''''''), z'0'''''')))
F(false, nil, .(y'''''', z'''''), .(nil, .(nil, .(.(y'''''''''''', z''''''''''''), z'0'''''''')))) -> DEL(.(.(y'''''', z'''''), .(nil, .(nil, .(.(y'''''''''''', z''''''''''''), z'0'''''''')))))


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

DEL(.(nil, .(.(y''', z'''), .(nil, .(nil, .(.(y'''''''''''''', z''''''''''''''), z'0'''''''''')))))) -> F(false, nil, .(y''', z'''), .(nil, .(nil, .(.(y'''''''''''''', z''''''''''''''), z'0''''''''''))))
DEL(.(nil, .(nil, .(nil, z''''')))) -> F(true, nil, nil, .(nil, z'''''))
DEL(.(nil, .(.(y''', z'''), .(nil, .(nil, .(nil, z''''''''''''')))))) -> F(false, nil, .(y''', z'''), .(nil, .(nil, .(nil, z'''''''''''''))))
DEL(.(nil, .(.(y''', z'''), .(nil, .(.(y''''''''''', z''''''''''), z'0''''''))))) -> F(false, nil, .(y''', z'''), .(nil, .(.(y''''''''''', z''''''''''), z'0'''''')))
DEL(.(nil, .(nil, .(.(y'''''', z''''''), z'0'')))) -> F(true, nil, nil, .(.(y'''''', z''''''), z'0''))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(DEL(x1))=  x1  
  POL(false)=  0  
  POL(nil)=  1  
  POL(true)=  0  
  POL(.(x1, x2))=  x1 + x2  
  POL(F(x1, x2, x3, x4))=  x3 + x4  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 11
                 ↳Dependency Graph


Dependency Pairs:

F(true, nil, nil, .(nil, z''')) -> DEL(.(nil, .(nil, z''')))
F(false, .(x'''', y''''), nil, .(nil, .(nil, z'''''''))) -> DEL(.(nil, .(nil, .(nil, z'''''''))))
DEL(.(.(x''', y'''), .(nil, .(nil, .(nil, z'''''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(nil, z''''''''')))
F(false, nil, .(y'''''', z'''''), .(nil, .(nil, .(nil, z''''''''''')))) -> DEL(.(.(y'''''', z'''''), .(nil, .(nil, .(nil, z''''''''''')))))
F(false, .(x'''', y''''), nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
DEL(.(.(x''', y'''), .(nil, .(.(y''''''', z''''''), z'0'')))) -> F(false, .(x''', y'''), nil, .(.(y''''''', z''''''), z'0''))
F(false, nil, .(y'''''', z'''''), .(nil, .(.(y''''''''', z''''''''), z'0''''))) -> DEL(.(.(y'''''', z'''''), .(nil, .(.(y''''''''', z''''''''), z'0''''))))
F(true, nil, nil, .(.(y'''', z''''), z'0)) -> DEL(.(nil, .(.(y'''', z''''), z'0)))
F(false, .(x'''', y''''), nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))) -> DEL(.(nil, .(nil, .(.(y'''''''', z''''''''), z'0''''))))
DEL(.(.(x''', y'''), .(nil, .(nil, .(.(y'''''''''', z''''''''''), z'0''''''))))) -> F(false, .(x''', y'''), nil, .(nil, .(.(y'''''''''', z''''''''''), z'0'''''')))
F(false, nil, .(y'''''', z'''''), .(nil, .(nil, .(.(y'''''''''''', z''''''''''''), z'0'''''''')))) -> DEL(.(.(y'''''', z'''''), .(nil, .(nil, .(.(y'''''''''''', z''''''''''''), z'0'''''''')))))


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes