Termination Proof using AProVE

Term Rewriting System R:
[x, y]
f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(s(x)) -> F(x)
G(s(x), s(y)) -> IF(f(x), s(x), s(y))
G(s(x), s(y)) -> F(x)
G(x, c(y)) -> G(x, g(s(c(y)), y))
G(x, c(y)) -> G(s(c(y)), 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:
F(s(x)) -> F(x)

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))
Dependency Pairs:
G(x, c(y)) -> G(s(c(y)), y)
G(x, c(y)) -> G(x, g(s(c(y)), y))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
F(s(x)) -> F(x)

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))
Dependency Pairs:
G(x, c(y)) -> G(s(c(y)), y)
G(x, c(y)) -> G(x, g(s(c(y)), y))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

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