Termination Proof using AProVE

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))

Termination of R to be shown.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
.'(.(x, y), z) -> .'(y, z)
.'(.(x, y), z) -> .'(x, .(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))
Dependency Pairs:
I(.(x, y)) -> I(x)
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))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
.'(.(x, y), z) -> .'(y, z)
.'(.(x, y), z) -> .'(x, .(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))
Dependency Pairs:
I(.(x, y)) -> I(x)
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))

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