Termination Proof using AProVE

Term Rewriting System R:
[X, Y]
+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> Y

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(X, s(Y)) -> +'(X, Y)
DOUBLE(X) -> +'(X, X)
F(0, s(0), X) -> F(X, double(X), X)
F(0, s(0), X) -> DOUBLE(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:
+'(X, s(Y)) -> +'(X, Y)

Rules:

+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> Y
Dependency Pair:
F(0, s(0), X) -> F(X, double(X), X)

Rules:

+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> 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:

+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> Y
Dependency Pair:
F(0, s(0), X) -> F(X, double(X), X)

Rules:

+(X, 0) -> X
+(X, s(Y)) -> s(+(X, Y))
double(X) -> +(X, X)
f(0, s(0), X) -> f(X, double(X), X)
g(X, Y) -> X
g(X, Y) -> Y

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