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

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Inst

       →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)))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
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 w.r.t. to the implicit AFS 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.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳Inst

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Instantiation Transformation

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

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))

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Inst

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
FROM(s(X'')) -> FROM(s(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

         ↳Polo

       →DP Problem 2

         ↳Inst

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
FROM(s(X'')) -> FROM(s(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

         ↳Polo

       →DP Problem 2

         ↳Inst

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
FROM(s(X'')) -> FROM(s(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:
0:00 minutes

POL(FST(x₁, x₂))	= x₁
POL(cons(x₁, x₂))	= 0
POL(s(x₁))	= 1 + x₁