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:

f(x, y, w, w, a) → g1(x, x, y, w)
f(x, y, w, a, a) → g1(y, x, x, w)
f(x, y, a, a, w) → g2(x, y, y, w)
f(x, y, a, w, w) → g2(y, y, x, w)
g1(x, x, y, a) → h(x, y)
g1(y, x, x, a) → h(x, y)
g2(x, y, y, a) → h(x, y)
g2(y, y, x, a) → h(x, y)
h(x, x) → x

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive Path Order [RPO].
Precedence:

f₅ > [a, g1₄, g2₄] > h₂

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

f(x, y, w, w, a) → g1(x, x, y, w)
f(x, y, w, a, a) → g1(y, x, x, w)
f(x, y, a, a, w) → g2(x, y, y, w)
f(x, y, a, w, w) → g2(y, y, x, w)
g1(x, x, y, a) → h(x, y)
g1(y, x, x, a) → h(x, y)
g2(x, y, y, a) → h(x, y)
g2(y, y, x, a) → h(x, y)
h(x, x) → 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