Termination Proof using AProVE

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

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →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:
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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))
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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))
Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

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

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →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:
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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))
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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))
Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

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

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →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:
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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))
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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))
Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

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

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →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:
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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))
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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
len(nil) -> 0
len(cons(X, Z)) -> s(len(Z))
Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

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

Termination of R could not be shown.
Duration:
0:01 minutes