Termination Proof using AProVE

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))

Innermost Termination of R to be shown.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

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

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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

three new Dependency Pairs are created:

F(.(.(.(x'', y''), y0), z'')) -> F(.(.(x'', y''), .(y0, z'')))
F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))
F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))
F(.(nil, .(nil, y''))) -> F(.(nil, y''))
F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))
F(.(.(.(x'', y''), y0), z'')) -> F(.(.(x'', y''), .(y0, z'')))
F(.(nil, .(.(x'', y''), z''))) -> F(.(.(x'', y''), z''))

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))

Strategy:
innermost
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))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
F(.(.(nil, .(x'''', y'''')), z')) -> F(.(nil, .(.(x'''', y''''), z')))
F(.(nil, .(nil, y''))) -> F(.(nil, y''))
F(.(.(nil, nil), z')) -> F(.(nil, .(nil, z')))
F(.(.(.(x'', y''), y0), z'')) -> F(.(.(x'', y''), .(y0, z'')))
F(.(nil, .(.(x'', y''), z''))) -> F(.(.(x'', y''), z''))

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))

Strategy:
innermost
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))

Strategy:
innermost

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