Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z]
and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
Dependency Pair:
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
Dependency Pair:
FROM(X) -> FROM(s(X))

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
Dependency Pair:
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
Dependency Pair:
FROM(X) -> FROM(s(X))

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
ADD(s(X), Y) -> ADD(X, Y)

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
Dependency Pair:
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
Dependency Pair:
FROM(X) -> FROM(s(X))

Rules:

and(true, X) -> X
and(false, Y) -> false
if(true, X, Y) -> X
if(false, X, Y) -> Y
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))

Termination of R could not be shown.
Duration:
0:00 minutes