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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Lexicographic path order with status [LPO].
Precedence:

f₂ > g1₃ > h₂
f₂ > g2₃ > h₂

Status:

g1₃: [1,2,3]
f₂: [1,2]
h₂: [1,2]
g2₃: [1,2,3]

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

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