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:

+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

f₂ > a > b > +₂

Status:

a: multiset
f₂: multiset
b: multiset
+₂: [1,2]

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

+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))

(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