Term Rewriting System R:
[x, y, z, k, l]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
MINUS(minus(x, y), z) -> PLUS(y, z)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
PLUS(s(x), y) -> PLUS(x, y)
APP(cons(x, l), k) -> APP(l, k)
SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))
SUM(cons(x, cons(y, l))) -> PLUS(x, y)
SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))
SUM(app(l, cons(x, cons(y, k)))) -> APP(l, sum(cons(x, cons(y, k))))
SUM(app(l, cons(x, cons(y, k)))) -> SUM(cons(x, cons(y, k)))

Furthermore, R contains six SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

PLUS(s(x), y) -> PLUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pair can be strictly oriented:

PLUS(s(x), y) -> PLUS(x, y)

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(PLUS(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 7`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

APP(cons(x, l), k) -> APP(l, k)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pair can be strictly oriented:

APP(cons(x, l), k) -> APP(l, k)

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(cons(x1, x2)) =  1 + x2 POL(APP(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 8`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pairs:

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
MINUS(s(x), s(y)) -> MINUS(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(x, y)

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  0 POL(0) =  0 POL(MINUS(x1, x2)) =  x1 POL(minus(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 9`
`             ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pair can be strictly oriented:

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  0 POL(0) =  0 POL(MINUS(x1, x2)) =  x1 POL(minus(x1, x2)) =  1 + x1 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 9`
`             ↳Polo`
`             ...`
`               →DP Problem 10`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polynomial Ordering`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pair can be strictly oriented:

SUM(cons(x, cons(y, l))) -> SUM(cons(plus(x, y), l))

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  0 POL(0) =  0 POL(SUM(x1)) =  x1 POL(cons(x1, x2)) =  1 + x2 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`           →DP Problem 11`
`             ↳Dependency Graph`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polynomial Ordering`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pair can be strictly oriented:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

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

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  0 POL(QUOT(x1, x2)) =  x1 POL(0) =  1 POL(minus(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 12`
`             ↳Dependency Graph`
`       →DP Problem 6`
`         ↳Nar`

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Narrowing Transformation`

Dependency Pair:

SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

SUM(app(l, cons(x, cons(y, k)))) -> SUM(app(l, sum(cons(x, cons(y, k)))))
three new Dependency Pairs are created:

SUM(app(nil, cons(x', cons(y', k'')))) -> SUM(sum(cons(x', cons(y', k''))))
SUM(app(cons(x'', l''), cons(x0, cons(y', k'')))) -> SUM(cons(x'', app(l'', sum(cons(x0, cons(y', k''))))))
SUM(app(l, cons(x'', cons(y'', k')))) -> SUM(app(l, sum(cons(plus(x'', y''), k'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Narrowing Transformation`

Dependency Pairs:

SUM(app(l, cons(x'', cons(y'', k')))) -> SUM(app(l, sum(cons(plus(x'', y''), k'))))
SUM(app(nil, cons(x', cons(y', k'')))) -> SUM(sum(cons(x', cons(y', k''))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

SUM(app(nil, cons(x', cons(y', k'')))) -> SUM(sum(cons(x', cons(y', k''))))
one new Dependency Pair is created:

SUM(app(nil, cons(x'', cons(y'', k''')))) -> SUM(sum(cons(plus(x'', y''), k''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 14`
`                 ↳Narrowing Transformation`

Dependency Pairs:

SUM(app(nil, cons(x'', cons(y'', k''')))) -> SUM(sum(cons(plus(x'', y''), k''')))
SUM(app(l, cons(x'', cons(y'', k')))) -> SUM(app(l, sum(cons(plus(x'', y''), k'))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

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

SUM(app(l, cons(x'', cons(y'', k')))) -> SUM(app(l, sum(cons(plus(x'', y''), k'))))
six new Dependency Pairs are created:

SUM(app(nil, cons(x''', cons(y''', k'')))) -> SUM(sum(cons(plus(x''', y'''), k'')))
SUM(app(cons(x', l''), cons(x''', cons(y''', k'')))) -> SUM(cons(x', app(l'', sum(cons(plus(x''', y'''), k'')))))
SUM(app(l, cons(x''', cons(y''', nil)))) -> SUM(app(l, cons(plus(x''', y'''), nil)))
SUM(app(l, cons(x''', cons(y''', cons(y', l''))))) -> SUM(app(l, sum(cons(plus(plus(x''', y'''), y'), l''))))
SUM(app(l, cons(0, cons(y''', k')))) -> SUM(app(l, sum(cons(y''', k'))))
SUM(app(l, cons(s(x'), cons(y''', k')))) -> SUM(app(l, sum(cons(s(plus(x', y''')), k'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Polynomial Ordering`

Dependency Pairs:

SUM(app(l, cons(s(x'), cons(y''', k')))) -> SUM(app(l, sum(cons(s(plus(x', y''')), k'))))
SUM(app(l, cons(0, cons(y''', k')))) -> SUM(app(l, sum(cons(y''', k'))))
SUM(app(l, cons(x''', cons(y''', cons(y', l''))))) -> SUM(app(l, sum(cons(plus(plus(x''', y'''), y'), l''))))
SUM(app(l, cons(x''', cons(y''', nil)))) -> SUM(app(l, cons(plus(x''', y'''), nil)))
SUM(app(nil, cons(x''', cons(y''', k'')))) -> SUM(sum(cons(plus(x''', y'''), k'')))
SUM(app(nil, cons(x'', cons(y'', k''')))) -> SUM(sum(cons(plus(x'', y''), k''')))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pairs can be strictly oriented:

SUM(app(nil, cons(x''', cons(y''', k'')))) -> SUM(sum(cons(plus(x''', y'''), k'')))
SUM(app(nil, cons(x'', cons(y'', k''')))) -> SUM(sum(cons(plus(x'', y''), k''')))

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

sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  0 POL(0) =  0 POL(SUM(x1)) =  1 + x1 POL(cons(x1, x2)) =  0 POL(nil) =  1 POL(s(x1)) =  0 POL(sum(x1)) =  0 POL(app(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 16`
`                 ↳Polynomial Ordering`

Dependency Pairs:

SUM(app(l, cons(s(x'), cons(y''', k')))) -> SUM(app(l, sum(cons(s(plus(x', y''')), k'))))
SUM(app(l, cons(0, cons(y''', k')))) -> SUM(app(l, sum(cons(y''', k'))))
SUM(app(l, cons(x''', cons(y''', cons(y', l''))))) -> SUM(app(l, sum(cons(plus(plus(x''', y'''), y'), l''))))
SUM(app(l, cons(x''', cons(y''', nil)))) -> SUM(app(l, cons(plus(x''', y'''), nil)))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pair can be strictly oriented:

SUM(app(l, cons(0, cons(y''', k')))) -> SUM(app(l, sum(cons(y''', k'))))

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

plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  x2 POL(0) =  1 POL(SUM(x1)) =  1 + x1 POL(cons(x1, x2)) =  x1 + x2 POL(nil) =  0 POL(s(x1)) =  0 POL(sum(x1)) =  x1 POL(app(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 17`
`                 ↳Polynomial Ordering`

Dependency Pairs:

SUM(app(l, cons(s(x'), cons(y''', k')))) -> SUM(app(l, sum(cons(s(plus(x', y''')), k'))))
SUM(app(l, cons(x''', cons(y''', cons(y', l''))))) -> SUM(app(l, sum(cons(plus(plus(x''', y'''), y'), l''))))
SUM(app(l, cons(x''', cons(y''', nil)))) -> SUM(app(l, cons(plus(x''', y'''), nil)))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

The following dependency pairs can be strictly oriented:

SUM(app(l, cons(s(x'), cons(y''', k')))) -> SUM(app(l, sum(cons(s(plus(x', y''')), k'))))
SUM(app(l, cons(x''', cons(y''', cons(y', l''))))) -> SUM(app(l, sum(cons(plus(plus(x''', y'''), y'), l''))))
SUM(app(l, cons(x''', cons(y''', nil)))) -> SUM(app(l, cons(plus(x''', y'''), nil)))

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

sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  0 POL(0) =  1 POL(SUM(x1)) =  x1 POL(cons(x1, x2)) =  1 + x2 POL(nil) =  0 POL(s(x1)) =  0 POL(sum(x1)) =  1 POL(app(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 18`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
app(nil, k) -> k
app(l, nil) -> l
app(cons(x, l), k) -> cons(x, app(l, k))
sum(cons(x, nil)) -> cons(x, nil)
sum(cons(x, cons(y, l))) -> sum(cons(plus(x, y), l))
sum(app(l, cons(x, cons(y, k)))) -> sum(app(l, sum(cons(x, cons(y, k)))))

Using the Dependency Graph resulted in no new DP problems.

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