Termination Proof using AProVE

Term Rewriting System R:
[x, y]
f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(s(x)) -> G(x, s(x))
G(s(x), y) -> G(x, +(y, s(x)))
G(s(x), y) -> +'(y, s(x))
G(s(x), y) -> G(x, s(+(y, x)))
G(s(x), y) -> +'(y, x)
+'(x, s(y)) -> +'(x, 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:
+'(x, s(y)) -> +'(x, y)

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pairs:
G(s(x), y) -> G(x, s(+(y, x)))
G(s(x), y) -> G(x, +(y, s(x)))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pairs:
G(s(x), y) -> G(x, s(+(y, x)))
G(s(x), y) -> G(x, +(y, s(x)))

Rules:

f(0) -> 1
f(s(x)) -> g(x, s(x))
g(0, y) -> y
g(s(x), y) -> g(x, +(y, s(x)))
g(s(x), y) -> g(x, s(+(y, x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

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