Term Rewriting System R:
[k, l, 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)))))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains four SCCs.

R
DPs
→DP Problem 1
Polynomial Ordering
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement 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 5
Dependency Graph
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo

Dependency Pair:

Rules:

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

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

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

There are no usable rules using the Ce-refinement 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 2
Polo
→DP Problem 6
Dependency Graph
→DP Problem 3
Polo
→DP Problem 4
Polo

Dependency Pair:

Rules:

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

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

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules using the Ce-refinement can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  x2 POL(0) =  0 POL(SUM(x1)) =  1 + 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 7
Dependency Graph
→DP Problem 4
Polo

Dependency Pair:

Rules:

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

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

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules using the Ce-refinement can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(plus(x1, x2)) =  x2 POL(0) =  0 POL(SUM(x1)) =  x1 POL(cons(x1, x2)) =  1 + x2 POL(nil) =  0 POL(sum(x1)) =  1 POL(s(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 8
Dependency Graph

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

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