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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(b) =  0 POL(a) =  1 POL(f(x1, x2)) =  x1 POL(F(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳Polynomial Ordering`

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(b) =  0 POL(a) =  1 POL(f(x1, x2)) =  x1 + x2 POL(F(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳Polo`
`             ...`
`               →DP Problem 3`
`                 ↳Narrowing Transformation`

Dependency Pairs:

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

Rules:

f(f(a, a), x) -> f(a, 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, a), x) -> F(a, f(b, f(a, x)))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳Polo`
`             ...`
`               →DP Problem 4`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

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

Rules:

f(f(a, a), x) -> f(a, 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)
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳Polo`
`             ...`
`               →DP Problem 5`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

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

Rules:

f(f(a, a), x) -> f(a, 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:
0:00 minutes