Term Rewriting System R:
[n, x, m, y]
sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))
SUM(cons(0, x), y) -> SUM(x, y)
WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
WEIGHT(cons(n, cons(m, x))) -> SUM(cons(n, cons(m, x)), cons(0, x))

Furthermore, R contains two SCCs.

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

Dependency Pairs:

SUM(cons(0, x), y) -> SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

The following dependency pair can be strictly oriented:

SUM(cons(0, x), y) -> SUM(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(SUM(x1, x2)) =  x1 POL(0) =  0 POL(cons(x1, x2)) =  1 + x2 POL(s(x1)) =  0

resulting in one new DP problem.

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

Dependency Pair:

SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

The following dependency pair can be strictly oriented:

SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), 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(SUM(x1, x2)) =  1 + x1 POL(cons(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 3
Polo
...
→DP Problem 4
Dependency Graph
→DP Problem 2
Polo

Dependency Pair:

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

The following dependency pair can be strictly oriented:

WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))

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

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y

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

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 5
Dependency Graph

Dependency Pair:

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Using the Dependency Graph resulted in no new DP problems.

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