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))

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`
`         ↳Negative Polynomial Order`

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))

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

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

There are no usable rules (regarding the implicit AFS).
Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, ..., x3) ) = x3

POL( g(x1) ) = x1 + 1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳Semantic Labelling`

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))

Using Semantic Labelling, the DP problem could be transformed. The following model was found:
 f(x0, x1, x2) =  1 g(x0) =  x0 F(x0, x1, x2) =  1 + x2 0 =  0 1 =  1

From the dependency graph we obtain 8 (labeled) SCCs which each result in correspondingDP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳SemLab`
`             ...`
`               →DP Problem 3`
`                 ↳Modular Removal of Rules`

Dependency Pairs:

F111(g1(x), y, z) -> F111(x, y, z)
F111(x, g1(y), z) -> F111(x, y, z)

Rules:

f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(F_111(x1, x2, x3)) =  x1 + x2 + x3 POL(g_1(x1)) =  x1

We have the following set D of usable symbols: {F111}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F111(g1(x), y, z) -> F111(x, y, z)
F111(x, g1(y), z) -> F111(x, y, z)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳SemLab`
`             ...`
`               →DP Problem 4`
`                 ↳Modular Removal of Rules`

Dependency Pairs:

F100(g1(x), y, z) -> F100(x, y, z)
F100(x, g0(y), z) -> F100(x, y, z)

Rules:

f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(F_100(x1, x2, x3)) =  x1 + x2 + x3 POL(g_1(x1)) =  x1 POL(g_0(x1)) =  x1

We have the following set D of usable symbols: {F100}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F100(g1(x), y, z) -> F100(x, y, z)
F100(x, g0(y), z) -> F100(x, y, z)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳SemLab`
`             ...`
`               →DP Problem 5`
`                 ↳Modular Removal of Rules`

Dependency Pairs:

F011(g0(x), y, z) -> F011(x, y, z)
F011(x, g1(y), z) -> F011(x, y, z)

Rules:

f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(g_1(x1)) =  x1 POL(F_011(x1, x2, x3)) =  x1 + x2 + x3 POL(g_0(x1)) =  x1

We have the following set D of usable symbols: {F011}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F011(g0(x), y, z) -> F011(x, y, z)
F011(x, g1(y), z) -> F011(x, y, z)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳SemLab`
`             ...`
`               →DP Problem 6`
`                 ↳Modular Removal of Rules`

Dependency Pairs:

F110(g1(x), y, z) -> F110(x, y, z)
F110(x, g1(y), z) -> F110(x, y, z)

Rules:

f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(g_1(x1)) =  x1 POL(F_110(x1, x2, x3)) =  x1 + x2 + x3

We have the following set D of usable symbols: {F110}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F110(g1(x), y, z) -> F110(x, y, z)
F110(x, g1(y), z) -> F110(x, y, z)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳SemLab`
`             ...`
`               →DP Problem 7`
`                 ↳Modular Removal of Rules`

Dependency Pairs:

F001(g0(x), y, z) -> F001(x, y, z)
F001(x, g0(y), z) -> F001(x, y, z)

Rules:

f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(F_001(x1, x2, x3)) =  x1 + x2 + x3 POL(g_0(x1)) =  x1

We have the following set D of usable symbols: {F001}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F001(g0(x), y, z) -> F001(x, y, z)
F001(x, g0(y), z) -> F001(x, y, z)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳SemLab`
`             ...`
`               →DP Problem 8`
`                 ↳Modular Removal of Rules`

Dependency Pairs:

F000(g0(x), y, z) -> F000(x, y, z)
F000(x, g0(y), z) -> F000(x, y, z)

Rules:

f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(F_000(x1, x2, x3)) =  x1 + x2 + x3 POL(g_0(x1)) =  x1

We have the following set D of usable symbols: {F000}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F000(g0(x), y, z) -> F000(x, y, z)
F000(x, g0(y), z) -> F000(x, y, z)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳SemLab`
`             ...`
`               →DP Problem 9`
`                 ↳Modular Removal of Rules`

Dependency Pairs:

F101(g1(x), y, z) -> F101(x, y, z)
F101(x, g0(y), z) -> F101(x, y, z)

Rules:

f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(g_1(x1)) =  x1 POL(F_101(x1, x2, x3)) =  x1 + x2 + x3 POL(g_0(x1)) =  x1

We have the following set D of usable symbols: {F101}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F101(g1(x), y, z) -> F101(x, y, z)
F101(x, g0(y), z) -> F101(x, y, z)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Neg POLO`
`           →DP Problem 2`
`             ↳SemLab`
`             ...`
`               →DP Problem 10`
`                 ↳Modular Removal of Rules`

Dependency Pairs:

F010(g0(x), y, z) -> F010(x, y, z)
F010(x, g1(y), z) -> F010(x, y, z)

Rules:

f010(0, 1, x) -> f000(g0(x), g0(x), x)
f010(g0(x), y, z) -> g1(f010(x, y, z))
f010(x, g1(y), z) -> g1(f010(x, y, z))
f010(x, y, g0(z)) -> g1(f010(x, y, z))
f000(g0(x), y, z) -> g1(f000(x, y, z))
f000(x, g0(y), z) -> g1(f000(x, y, z))
f000(x, y, g0(z)) -> g1(f000(x, y, z))
f011(0, 1, x) -> f111(g1(x), g1(x), x)
f011(g0(x), y, z) -> g1(f011(x, y, z))
f011(x, g1(y), z) -> g1(f011(x, y, z))
f011(x, y, g1(z)) -> g1(f011(x, y, z))
f111(g1(x), y, z) -> g1(f111(x, y, z))
f111(x, g1(y), z) -> g1(f111(x, y, z))
f111(x, y, g1(z)) -> g1(f111(x, y, z))
f001(g0(x), y, z) -> g1(f001(x, y, z))
f001(x, g0(y), z) -> g1(f001(x, y, z))
f001(x, y, g1(z)) -> g1(f001(x, y, z))
f100(g1(x), y, z) -> g1(f100(x, y, z))
f100(x, g0(y), z) -> g1(f100(x, y, z))
f100(x, y, g0(z)) -> g1(f100(x, y, z))
f101(g1(x), y, z) -> g1(f101(x, y, z))
f101(x, g0(y), z) -> g1(f101(x, y, z))
f101(x, y, g1(z)) -> g1(f101(x, y, z))
f110(g1(x), y, z) -> g1(f110(x, y, z))
f110(x, g1(y), z) -> g1(f110(x, y, z))
f110(x, y, g0(z)) -> g1(f110(x, y, z))

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(F_010(x1, x2, x3)) =  x1 + x2 + x3 POL(g_1(x1)) =  x1 POL(g_0(x1)) =  x1

We have the following set D of usable symbols: {F010}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

F010(g0(x), y, z) -> F010(x, y, z)
F010(x, g1(y), z) -> F010(x, y, z)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

Termination of R successfully shown.
Duration:
0:07 minutes