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:

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

i₁ > 0 > +₂

Status:

i₁: [1]
0: multiset
+₂: [2,1]

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

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → 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