Term Rewriting System R:
[x, y, z]
.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))

Innermost 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)
I(.(x, y)) -> .'(i(y), i(x))
I(.(x, y)) -> I(y)
I(.(x, y)) -> I(x)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳AFS`

Dependency Pair:

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

Rules:

.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(.(x1, x2)) =  1 + x1 + x2 POL(.'(x1, x2)) =  x1 + x2

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 3`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳AFS`

Dependency Pair:

Rules:

.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳Argument Filtering and Ordering`

Dependency Pair:

I(.(x, y)) -> I(y)

Rules:

.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

I(.(x, y)) -> I(y)

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(I(x1)) =  x1 POL(.(x1, x2)) =  1 + x1 + x2

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

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

Dependency Pair:

Rules:

.(1, x) -> x
.(x, 1) -> x
.(i(x), x) -> 1
.(x, i(x)) -> 1
.(i(y), .(y, z)) -> z
.(y, .(i(y), z)) -> z
.(.(x, y), z) -> .(x, .(y, z))
i(1) -> 1
i(i(x)) -> x
i(.(x, y)) -> .(i(y), i(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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