Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Precedence:

i₁ > .₂
i₁ > 1

Status:

i₁: [1]
.₂: [1,2]
1: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(x))

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE