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

Innermost 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`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 2`
`         ↳UsableRules`

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

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 3`
`             ↳Size-Change Principle`
`       →DP Problem 2`
`         ↳UsableRules`

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))

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. SUM(cons(0, x), y) -> SUM(x, y)
2. SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1
2=2
{1, 2} , {1, 2}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1>1
{1, 2} , {1, 2}
1>1
2=2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)
s(x1) -> s(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳Usable Rules (Innermost)`

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

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Negative Polynomial Order`

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

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are 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 Order with Interpretation:

POL( WEIGHT(x1) ) = x1

POL( cons(x1, x2) ) = x2 + 1

POL( sum(x1, x2) ) = x2

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Neg 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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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