Term Rewriting System R:
[x, y, z]
f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

F(0, 1, x) -> F(g(x), g(x), x)
F(g(x), y, z) -> F(x, y, z)
F(x, g(y), z) -> F(x, y, z)
F(x, y, g(z)) -> F(x, y, z)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`

Dependency Pairs:

F(x, y, g(z)) -> F(x, y, z)
F(x, g(y), z) -> F(x, y, z)
F(g(x), y, z) -> F(x, y, z)
F(0, 1, x) -> F(g(x), g(x), x)

Rules:

f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(g(x1)) =  1 + x1

resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2, x3) -> x3
g(x1) -> g(x1)

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

The following remains to be proven:
Dependency Pairs:

F(x, g(y), z) -> F(x, y, z)
F(g(x), y, z) -> F(x, y, z)
F(0, 1, x) -> F(g(x), g(x), x)

Rules:

f(0, 1, x) -> f(g(x), g(x), x)
f(g(x), y, z) -> g(f(x, y, z))
f(x, g(y), z) -> g(f(x, y, z))
f(x, y, g(z)) -> g(f(x, y, z))

Strategy:

innermost

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