Termination Proof using AProVE

Term Rewriting System R:
[y, x]
int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> int_list(int(x, y))
int_list(nil) -> nil
int_list(.(x, y)) -> .(s(x), int_list(y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

INT(0, s(y)) -> INT(s(0), s(y))
INT(s(x), s(y)) -> INT_LIST(int(x, y))
INT(s(x), s(y)) -> INT(x, y)
INT_LIST(.(x, y)) -> INT_LIST(y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
INT_LIST(.(x, y)) -> INT_LIST(y)

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> int_list(int(x, y))
int_list(nil) -> nil
int_list(.(x, y)) -> .(s(x), int_list(y))
Dependency Pairs:
INT(s(x), s(y)) -> INT(x, y)
INT(0, s(y)) -> INT(s(0), s(y))

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> int_list(int(x, y))
int_list(nil) -> nil
int_list(.(x, y)) -> .(s(x), int_list(y))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
INT_LIST(.(x, y)) -> INT_LIST(y)

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> int_list(int(x, y))
int_list(nil) -> nil
int_list(.(x, y)) -> .(s(x), int_list(y))
Dependency Pairs:
INT(s(x), s(y)) -> INT(x, y)
INT(0, s(y)) -> INT(s(0), s(y))

Rules:

int(0, 0) -> .(0, nil)
int(0, s(y)) -> .(0, int(s(0), s(y)))
int(s(x), 0) -> nil
int(s(x), s(y)) -> int_list(int(x, y))
int_list(nil) -> nil
int_list(.(x, y)) -> .(s(x), int_list(y))

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