Termination Proof using AProVE

Term Rewriting System R:
[f, g, x, l, xs]
app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs
last -> app(app(compose, hd), reverse)
init -> app(app(compose, reverse), app(app(compose, tl), reverse))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(app(compose, f), g), x) -> APP(g, app(f, x))
APP(app(app(compose, f), g), x) -> APP(f, x)
APP(reverse, l) -> APP(app(reverse2, l), nil)
APP(reverse, l) -> APP(reverse2, l)
APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(reverse2, xs), app(app(cons, x), l))
APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(reverse2, xs)
APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(cons, x), l)
LAST -> APP(app(compose, hd), reverse)
LAST -> APP(compose, hd)
INIT -> APP(app(compose, reverse), app(app(compose, tl), reverse))
INIT -> APP(compose, reverse)
INIT -> APP(app(compose, tl), reverse)
INIT -> APP(compose, tl)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(cons, x), l)
APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(reverse2, xs), app(app(cons, x), l))
APP(reverse, l) -> APP(app(reverse2, l), nil)
APP(app(app(compose, f), g), x) -> APP(f, x)
APP(app(app(compose, f), g), x) -> APP(g, app(f, x))

Rules:

app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs
last -> app(app(compose, hd), reverse)
init -> app(app(compose, reverse), app(app(compose, tl), reverse))

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

APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(cons, x), l)

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Polynomial Ordering

Dependency Pairs:

APP(reverse, l) -> APP(app(reverse2, l), nil)
APP(app(app(compose, f), g), x) -> APP(f, x)
APP(app(app(compose, f), g), x) -> APP(g, app(f, x))
APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(reverse2, xs), app(app(cons, x), l))

Rules:

app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs
last -> app(app(compose, hd), reverse)
init -> app(app(compose, reverse), app(app(compose, tl), reverse))

The following dependency pairs can be strictly oriented:

APP(app(app(compose, f), g), x) -> APP(f, x)
APP(app(app(compose, f), g), x) -> APP(g, app(f, x))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(reverse) = 0

POL(cons) = 0

POL(hd) = 1

POL(nil) = 0

POL(tl) = 1

POL(compose) = 1

POL(app(x₁, x₂)) = x₁ + x₂

POL(APP(x₁, x₂)) = 1 + x₁ + x₂

POL(reverse2) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Polo

...

               →DP Problem 3

                 ↳Polynomial Ordering

Dependency Pairs:

APP(reverse, l) -> APP(app(reverse2, l), nil)
APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(reverse2, xs), app(app(cons, x), l))

Rules:

app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs
last -> app(app(compose, hd), reverse)
init -> app(app(compose, reverse), app(app(compose, tl), reverse))

The following dependency pair can be strictly oriented:

APP(reverse, l) -> APP(app(reverse2, l), nil)

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(reverse) = 1

POL(cons) = 0

POL(hd) = 1

POL(nil) = 0

POL(tl) = 1

POL(compose) = 0

POL(app(x₁, x₂)) = x₁ + x₂

POL(APP(x₁, x₂)) = 1 + x₁ + x₂

POL(reverse2) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Polo

...

               →DP Problem 4

                 ↳Polynomial Ordering

Dependency Pair:

APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(reverse2, xs), app(app(cons, x), l))

Rules:

app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs
last -> app(app(compose, hd), reverse)
init -> app(app(compose, reverse), app(app(compose, tl), reverse))

The following dependency pair can be strictly oriented:

APP(app(reverse2, app(app(cons, x), xs)), l) -> APP(app(reverse2, xs), app(app(cons, x), l))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(reverse) = 0

POL(cons) = 1

POL(hd) = 1

POL(nil) = 0

POL(tl) = 1

POL(compose) = 0

POL(app(x₁, x₂)) = x₁ + x₂

POL(APP(x₁, x₂)) = x₁

POL(reverse2) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Polo

...

               →DP Problem 5

                 ↳Dependency Graph

Dependency Pair:

Rules:

app(app(app(compose, f), g), x) -> app(g, app(f, x))
app(reverse, l) -> app(app(reverse2, l), nil)
app(app(reverse2, nil), l) -> l
app(app(reverse2, app(app(cons, x), xs)), l) -> app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) -> x
app(tl, app(app(cons, x), xs)) -> xs
last -> app(app(compose, hd), reverse)
init -> app(app(compose, reverse), app(app(compose, tl), reverse))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes

POL(reverse)	= 0
POL(cons)	= 0
POL(hd)	= 1
POL(nil)	= 0
POL(tl)	= 1
POL(compose)	= 1
POL(app(x₁, x₂))	= x₁ + x₂
POL(APP(x₁, x₂))	= 1 + x₁ + x₂
POL(reverse2)	= 0