Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
f(f(a, b), x) -> f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) -> f(f(x, y), z)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(f(a, b), x) -> F(b, f(a, f(c, f(b, f(a, x)))))
F(f(a, b), x) -> F(a, f(c, f(b, f(a, x))))
F(f(a, b), x) -> F(c, f(b, f(a, x)))
F(f(a, b), x) -> F(b, f(a, x))
F(f(a, b), x) -> F(a, x)
F(x, f(y, z)) -> F(f(x, y), z)
F(x, f(y, z)) -> F(x, y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

F(f(a, b), x) -> F(a, x)
F(f(a, b), x) -> F(b, f(a, x))
F(f(a, b), x) -> F(c, f(b, f(a, x)))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, f(c, f(b, f(a, x))))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), x) -> F(b, f(a, f(c, f(b, f(a, x)))))

Rules:

f(f(a, b), x) -> f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

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

F(f(a, b), x) -> F(b, f(a, f(c, f(b, f(a, x)))))

four new Dependency Pairs are created:

F(f(a, b), x'') -> F(b, f(f(a, c), f(b, f(a, x''))))
F(f(a, b), x'') -> F(b, f(a, f(f(c, b), f(a, x''))))
F(f(a, b), x'') -> F(b, f(a, f(c, f(f(b, a), x''))))
F(f(a, b), f(y', z')) -> F(b, f(a, f(c, f(b, f(f(a, y'), z')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

F(f(a, b), f(y', z')) -> F(b, f(a, f(c, f(b, f(f(a, y'), z')))))
F(f(a, b), x'') -> F(b, f(a, f(c, f(f(b, a), x''))))
F(f(a, b), x'') -> F(b, f(a, f(f(c, b), f(a, x''))))
F(f(a, b), x'') -> F(b, f(f(a, c), f(b, f(a, x''))))
F(f(a, b), x) -> F(b, f(a, x))
F(f(a, b), x) -> F(c, f(b, f(a, x)))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, f(c, f(b, f(a, x))))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), x) -> F(a, x)

Rules:

f(f(a, b), x) -> f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

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

F(f(a, b), x) -> F(a, f(c, f(b, f(a, x))))

three new Dependency Pairs are created:

F(f(a, b), x'') -> F(a, f(f(c, b), f(a, x'')))
F(f(a, b), x'') -> F(a, f(c, f(f(b, a), x'')))
F(f(a, b), f(y', z')) -> F(a, f(c, f(b, f(f(a, y'), z'))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

F(f(a, b), f(y', z')) -> F(a, f(c, f(b, f(f(a, y'), z'))))
F(f(a, b), x'') -> F(a, f(c, f(f(b, a), x'')))
F(f(a, b), x'') -> F(a, f(f(c, b), f(a, x'')))
F(f(a, b), x'') -> F(b, f(a, f(c, f(f(b, a), x''))))
F(f(a, b), x'') -> F(b, f(a, f(f(c, b), f(a, x''))))
F(f(a, b), x'') -> F(b, f(f(a, c), f(b, f(a, x''))))
F(f(a, b), x) -> F(a, x)
F(f(a, b), x) -> F(b, f(a, x))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(c, f(b, f(a, x)))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), f(y', z')) -> F(b, f(a, f(c, f(b, f(f(a, y'), z')))))

Rules:

f(f(a, b), x) -> f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

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

F(f(a, b), x) -> F(c, f(b, f(a, x)))

two new Dependency Pairs are created:

F(f(a, b), x'') -> F(c, f(f(b, a), x''))
F(f(a, b), f(y', z')) -> F(c, f(b, f(f(a, y'), z')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(a, b), f(y', z')) -> F(c, f(b, f(f(a, y'), z')))
F(f(a, b), x'') -> F(c, f(f(b, a), x''))
F(f(a, b), x'') -> F(a, f(c, f(f(b, a), x'')))
F(f(a, b), x'') -> F(a, f(f(c, b), f(a, x'')))
F(f(a, b), f(y', z')) -> F(b, f(a, f(c, f(b, f(f(a, y'), z')))))
F(f(a, b), x'') -> F(b, f(a, f(c, f(f(b, a), x''))))
F(f(a, b), x'') -> F(b, f(a, f(f(c, b), f(a, x''))))
F(f(a, b), x'') -> F(b, f(f(a, c), f(b, f(a, x''))))
F(f(a, b), x) -> F(a, x)
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(b, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), f(y', z')) -> F(a, f(c, f(b, f(f(a, y'), z'))))

Rules:

f(f(a, b), x) -> f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) -> f(f(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(f(a, b), x) -> F(a, x)

one new Dependency Pair is created:

F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
F(f(a, b), x'') -> F(c, f(f(b, a), x''))
F(f(a, b), f(y', z')) -> F(a, f(c, f(b, f(f(a, y'), z'))))
F(f(a, b), x'') -> F(a, f(c, f(f(b, a), x'')))
F(f(a, b), x'') -> F(a, f(f(c, b), f(a, x'')))
F(f(a, b), f(y', z')) -> F(b, f(a, f(c, f(b, f(f(a, y'), z')))))
F(f(a, b), x'') -> F(b, f(a, f(c, f(f(b, a), x''))))
F(f(a, b), x'') -> F(b, f(a, f(f(c, b), f(a, x''))))
F(f(a, b), x'') -> F(b, f(f(a, c), f(b, f(a, x''))))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(b, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), f(y', z')) -> F(c, f(b, f(f(a, y'), z')))

Rules:

f(f(a, b), x) -> f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) -> f(f(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, f(y, z)) -> F(x, y)

four new Dependency Pairs are created:

F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(a, b), f(y', z)) -> F(f(a, b), y')
F(f(a, b), f(f(y''', z'''), z)) -> F(f(a, b), f(y''', z'''))
F(f(a, b), f(f(y'''', z''''), z)) -> F(f(a, b), f(y'''', z''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

F(f(a, b), f(f(y'''', z''''), z)) -> F(f(a, b), f(y'''', z''''))
F(f(a, b), f(f(y''', z'''), z)) -> F(f(a, b), f(y''', z'''))
F(f(a, b), f(y', z)) -> F(f(a, b), y')
F(f(a, b), f(y', z')) -> F(c, f(b, f(f(a, y'), z')))
F(f(a, b), x'') -> F(c, f(f(b, a), x''))
F(f(a, b), f(y', z')) -> F(a, f(c, f(b, f(f(a, y'), z'))))
F(f(a, b), x'') -> F(a, f(c, f(f(b, a), x'')))
F(f(a, b), x'') -> F(a, f(f(c, b), f(a, x'')))
F(f(a, b), f(y', z')) -> F(b, f(a, f(c, f(b, f(f(a, y'), z')))))
F(f(a, b), x'') -> F(b, f(a, f(c, f(f(b, a), x''))))
F(f(a, b), x'') -> F(b, f(a, f(f(c, b), f(a, x''))))
F(f(a, b), x'') -> F(b, f(f(a, c), f(b, f(a, x''))))
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(a, b), x) -> F(b, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))

Rules:

f(f(a, b), x) -> f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) -> f(f(x, y), z)

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:02 minutes