Termination Proof using AProVE

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

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

NORM(g(x, y)) -> NORM(x)
F(x, g(y, z)) -> F(x, y)
REM(g(x, y), s(z)) -> REM(x, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
NORM(g(x, y)) -> NORM(x)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)
Dependency Pair:
F(x, g(y, z)) -> F(x, y)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)
Dependency Pair:
REM(g(x, y), s(z)) -> REM(x, z)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
NORM(g(x, y)) -> NORM(x)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)
Dependency Pair:
F(x, g(y, z)) -> F(x, y)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)
Dependency Pair:
REM(g(x, y), s(z)) -> REM(x, z)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
NORM(g(x, y)) -> NORM(x)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)
Dependency Pair:
F(x, g(y, z)) -> F(x, y)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)
Dependency Pair:
REM(g(x, y), s(z)) -> REM(x, z)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

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