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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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