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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(f(f(a, b), c), x) -> F(b, f(a, f(c, f(b, x))))
F(f(f(a, b), c), x) -> F(a, f(c, f(b, x)))
F(f(f(a, b), c), x) -> F(c, f(b, x))
F(f(f(a, b), c), x) -> F(b, 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`
`         ↳Modular Removal of Rules`

Dependency Pairs:

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

Rules:

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

We have the following set of usable rules:

f(x, f(y, z)) -> f(f(x, y), z)
f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x))))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(c) =  0 POL(b) =  0 POL(a) =  1 POL(F(x1, x2)) =  1 + x1 + x2 POL(f(x1, x2)) =  x1 + x2

We have the following set D of usable symbols: {c, b, a, F, f}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

The result of this processor delivers one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳MRR`
`           →DP Problem 2`
`             ↳Negative Polynomial Order`

Dependency Pairs:

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

Rules:

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

The following Dependency Pair can be strictly oriented using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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

Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, x2) ) = x1

POL( f(x1, x2) ) = x1

POL( a ) = 1

POL( b ) = 0

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳MRR`
`           →DP Problem 2`
`             ↳Neg POLO`
`             ...`
`               →DP Problem 3`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

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

Rules:

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

Termination of R could not be shown.
Duration:
0:41 minutes