Termination Proof using AProVE

Term Rewriting System R:
[x, y]
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
F(s(x)) -> -'(s(x), g(f(x)))
F(s(x)) -> G(f(x))
F(s(x)) -> F(x)
G(s(x)) -> -'(s(x), f(g(x)))
G(s(x)) -> F(g(x))
G(s(x)) -> G(x)

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:
-'(s(x), s(y)) -> -'(x, y)

Rules:

-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))
Dependency Pairs:
G(s(x)) -> G(x)
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))
F(s(x)) -> G(f(x))

Rules:

-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
-'(s(x), s(y)) -> -'(x, y)

Rules:

-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))
Dependency Pairs:
G(s(x)) -> G(x)
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))
F(s(x)) -> G(f(x))

Rules:

-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
f(0) -> 0
f(s(x)) -> -(s(x), g(f(x)))
g(0) -> s(0)
g(s(x)) -> -(s(x), f(g(x)))

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