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(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(x), y)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

j₂ > f₂ > h2₃ > s₁
j₂ > g₁ > h2₃ > s₁
j₂ > k₁ > h1₂ > h2₃ > s₁
j₂ > k₁ > h1₂ > 0
h₁ > h1₂ > h2₃ > s₁
h₁ > h1₂ > 0

Status:

i₁: [1]
j₂: [1,2]
f₂: [2,1]
g₁: [1]
k₁: [1]
h1₂: [2,1]
h2₃: [2,1,3]
h₁: [1]
s₁: [1]
0: []

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

f(j(x, y), y) → g(f(x, k(y)))
f(x, h1(y, z)) → h2(0, x, h1(y, z))
g(h2(x, y, h1(z, u))) → h2(s(x), y, h1(z, u))
h2(x, j(y, h1(z, u)), h1(z, u)) → h2(s(x), y, h1(s(z), u))
i(f(x, h(y))) → y
i(h2(s(x), y, h1(x, z))) → z
k(h(x)) → h1(0, x)
k(h1(x, y)) → h1(s(x), y)

(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