Term Rewriting System R:
[Z, X, Y]
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

FST(s(X), cons(Y, Z)) -> FST(X, Z)
FROM(X) -> FROM(s(X))
LEN(cons(X, Z)) -> LEN(Z)

Furthermore, R contains four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Remaining`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

FST(s(X), cons(Y, Z)) -> FST(X, Z)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

The following dependency pair can be strictly oriented:

FST(s(X), cons(Y, Z)) -> FST(X, Z)

There are no usable rules using the Ce-refinement that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 5`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Remaining`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`

Dependency Pair:

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

• Dependency Pair:

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

• Dependency Pair:

LEN(cons(X, Z)) -> LEN(Z)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

• Dependency Pair:

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

• Dependency Pair:

LEN(cons(X, Z)) -> LEN(Z)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

• Dependency Pair:

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))

• Dependency Pair:

LEN(cons(X, Z)) -> LEN(Z)

Rules:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, fst(X, Z))
from(X) -> cons(X, from(s(X)))