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`
`         ↳Polynomial 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 implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(g(x1)) =  1 + x1 POL(1) =  0 POL(F(x1, x2, x3)) =  x3

resulting in one new DP problem.

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

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

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

F(g(x), y, z) -> F(x, y, z)
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 3`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

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

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

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

F(x, g(y), z) -> F(x, y, z)
five new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

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

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:
0:01 minutes