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

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
+'(f(x), f(y)) -> +'(x, y)
+'(f(x), +(f(y), z)) -> +'(f(+(x, y)), z)
+'(f(x), +(f(y), z)) -> +'(x, y)

Furthermore, R contains one SCC.

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

Dependency Pairs:

+'(f(x), +(f(y), z)) -> +'(x, y)
+'(f(x), +(f(y), z)) -> +'(f(+(x, y)), z)
+'(f(x), f(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))

Rules:

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)

The following dependency pairs can be strictly oriented:

+'(f(x), +(f(y), z)) -> +'(x, y)
+'(+(x, y), z) -> +'(y, z)

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

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(+(x1, x2)) =  1 + x1 + x2 POL(f(x1)) =  x1 POL(+'(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

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

Dependency Pairs:

+'(f(x), +(f(y), z)) -> +'(f(+(x, y)), z)
+'(f(x), f(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))

Rules:

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)

The following dependency pairs can be strictly oriented:

+'(f(x), +(f(y), z)) -> +'(f(+(x, y)), z)
+'(f(x), f(y)) -> +'(x, y)

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

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(+(x1, x2)) =  x1 + x2 POL(f(x1)) =  1 + x1 POL(+'(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳Polo`
`             ...`
`               →DP Problem 3`
`                 ↳Polynomial Ordering`

Dependency Pair:

+'(+(x, y), z) -> +'(x, +(y, z))

Rules:

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)

The following dependency pair can be strictly oriented:

+'(+(x, y), z) -> +'(x, +(y, z))

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 2`
`             ↳Polo`
`             ...`
`               →DP Problem 4`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

+(+(x, y), z) -> +(x, +(y, z))
+(f(x), f(y)) -> f(+(x, y))
+(f(x), +(f(y), z)) -> +(f(+(x, y)), z)

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes