Term Rewriting System R:
[y, x, z]
f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(.(nil, y)) -> F(y)
F(.(.(x, y), z)) -> F(.(x, .(y, z)))
G(.(x, nil)) -> G(x)
G(.(x, .(y, z))) -> G(.(.(x, y), z))

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pairs:

F(.(.(x, y), z)) -> F(.(x, .(y, z)))
F(.(nil, y)) -> F(y)

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

• Dependency Pairs:

G(.(x, .(y, z))) -> G(.(.(x, y), z))
G(.(x, nil)) -> G(x)

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
• Dependency Pairs:

F(.(.(x, y), z)) -> F(.(x, .(y, z)))
F(.(nil, y)) -> F(y)

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

• Dependency Pairs:

G(.(x, .(y, z))) -> G(.(.(x, y), z))
G(.(x, nil)) -> G(x)

Rules:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

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