Term Rewriting System R:
[Z, X, Y, X1, X2]
afst(0, Z) -> nil
afst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
afst(X1, X2) -> fst(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alen(nil) -> 0
alen(cons(X, Z)) -> s(len(Z))
alen(X) -> len(X)
mark(fst(X1, X2)) -> afst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(len(X)) -> alen(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

AFST(s(X), cons(Y, Z)) -> MARK(Y)
AFROM(X) -> MARK(X)
MARK(fst(X1, X2)) -> AFST(mark(X1), mark(X2))
MARK(fst(X1, X2)) -> MARK(X1)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(from(X)) -> AFROM(mark(X))
MARK(from(X)) -> MARK(X)
MARK(len(X)) -> ALEN(mark(X))
MARK(len(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Negative Polynomial Order`

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(len(X)) -> MARK(X)
MARK(from(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(fst(X1, X2)) -> AFST(mark(X1), mark(X2))
AFST(s(X), cons(Y, Z)) -> MARK(Y)

Rules:

afst(0, Z) -> nil
afst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
afst(X1, X2) -> fst(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alen(nil) -> 0
alen(cons(X, Z)) -> s(len(Z))
alen(X) -> len(X)
mark(fst(X1, X2)) -> afst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(len(X)) -> alen(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

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

MARK(cons(X1, X2)) -> MARK(X1)
MARK(from(X)) -> MARK(X)
MARK(from(X)) -> AFROM(mark(X))
AFST(s(X), cons(Y, Z)) -> MARK(Y)

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

mark(fst(X1, X2)) -> afst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(len(X)) -> alen(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
afst(0, Z) -> nil
afst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
afst(X1, X2) -> fst(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alen(nil) -> 0
alen(cons(X, Z)) -> s(len(Z))
alen(X) -> len(X)

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x1) ) = x1

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

POL( AFROM(x1) ) = x1

POL( len(x1) ) = x1

POL( AADD(x1, x2) ) = x2

POL( add(x1, x2) ) = x1 + x2

POL( mark(x1) ) = x1

POL( AFST(x1, x2) ) = x2

POL( fst(x1, x2) ) = x1 + x2

POL( from(x1) ) = x1 + 1

POL( afst(x1, x2) ) = x1 + x2

POL( afrom(x1) ) = x1 + 1

POL( aadd(x1, x2) ) = x1 + x2

POL( alen(x1) ) = x1

POL( 0 ) = 0

POL( s(x1) ) = 0

POL( nil ) = 0

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳Dependency Graph`

Dependency Pairs:

MARK(len(X)) -> MARK(X)
AFROM(X) -> MARK(X)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(fst(X1, X2)) -> AFST(mark(X1), mark(X2))

Rules:

afst(0, Z) -> nil
afst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
afst(X1, X2) -> fst(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alen(nil) -> 0
alen(cons(X, Z)) -> s(len(Z))
alen(X) -> len(X)
mark(fst(X1, X2)) -> afst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(len(X)) -> alen(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Using the Dependency Graph the DP problem was split into 1 DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳DGraph`
`             ...`
`               →DP Problem 3`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(len(X)) -> MARK(X)

Rules:

afst(0, Z) -> nil
afst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
afst(X1, X2) -> fst(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alen(nil) -> 0
alen(cons(X, Z)) -> s(len(Z))
alen(X) -> len(X)
mark(fst(X1, X2)) -> afst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(len(X)) -> alen(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

mark(fst(X1, X2)) -> afst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(len(X)) -> alen(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
afst(0, Z) -> nil
afst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
afst(X1, X2) -> fst(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alen(nil) -> 0
alen(cons(X, Z)) -> s(len(Z))
alen(X) -> len(X)

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x1) ) = x1

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

POL( AADD(x1, x2) ) = x2

POL( mark(x1) ) = x1

POL( len(x1) ) = x1

POL( fst(x1, x2) ) = x1 + x2

POL( afst(x1, x2) ) = x1 + x2

POL( from(x1) ) = 0

POL( afrom(x1) ) = 0

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

POL( alen(x1) ) = x1

POL( 0 ) = 0

POL( s(x1) ) = 0

POL( nil ) = 0

POL( cons(x1, x2) ) = 0

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳DGraph`
`             ...`
`               →DP Problem 4`
`                 ↳Dependency Graph`

Dependency Pairs:

MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(len(X)) -> MARK(X)

Rules:

afst(0, Z) -> nil
afst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
afst(X1, X2) -> fst(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alen(nil) -> 0
alen(cons(X, Z)) -> s(len(Z))
alen(X) -> len(X)
mark(fst(X1, X2)) -> afst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(len(X)) -> alen(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Using the Dependency Graph the DP problem was split into 1 DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳DGraph`
`             ...`
`               →DP Problem 5`
`                 ↳Size-Change Principle`

Dependency Pairs:

MARK(len(X)) -> MARK(X)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)

Rules:

afst(0, Z) -> nil
afst(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
afst(X1, X2) -> fst(X1, X2)
afrom(X) -> cons(mark(X), from(s(X)))
afrom(X) -> from(X)
alen(nil) -> 0
alen(cons(X, Z)) -> s(len(Z))
alen(X) -> len(X)
mark(fst(X1, X2)) -> afst(mark(X1), mark(X2))
mark(from(X)) -> afrom(mark(X))
mark(len(X)) -> alen(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

We number the DPs as follows:
1. MARK(len(X)) -> MARK(X)
2. MARK(fst(X1, X2)) -> MARK(X2)
3. MARK(fst(X1, X2)) -> MARK(X1)
and get the following Size-Change Graph(s):
{3, 2, 1} , {3, 2, 1}
1>1

which lead(s) to this/these maximal multigraph(s):
{3, 2, 1} , {3, 2, 1}
1>1

DP: empty set
Oriented Rules: none

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

with Argument Filtering System:
len(x1) -> len(x1)
fst(x1, x2) -> fst(x1, x2)

We obtain no new DP problems.

Termination of R successfully shown.
Duration:
0:02 minutes