Termination Proof using AProVE

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

FIB(s(s(x))) -> +'(fib(s(x)), fib(x))
FIB(s(s(x))) -> FIB(s(x))
FIB(s(s(x))) -> FIB(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:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pairs:
FIB(s(s(x))) -> FIB(x)
FIB(s(s(x))) -> FIB(s(x))

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(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:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pairs:
FIB(s(s(x))) -> FIB(x)
FIB(s(s(x))) -> FIB(s(x))

Rules:

fib(0) -> 0
fib(s(0)) -> s(0)
fib(s(s(x))) -> +(fib(s(x)), fib(x))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

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