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:

D(t) → s(h)
D(constant) → h
D(b(x, y)) → b(D(x), D(y))
D(c(x, y)) → b(c(y, D(x)), c(x, D(y)))
D(m(x, y)) → m(D(x), D(y))
D(opp(x)) → opp(D(x))
D(div(x, y)) → m(div(D(x), y), div(c(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → b(c(c(y, pow(x, m(y, 1))), D(x)), c(c(pow(x, y), ln(x)), D(y)))
b(h, x) → x
b(x, h) → x
b(s(x), s(y)) → s(s(b(x, y)))
b(b(x, y), z) → b(x, b(y, z))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

D₁ > h > s₁
D₁ > c₂ > b₂ > s₁
D₁ > m₂ > s₁
D₁ > opp₁ > s₁
D₁ > pow₂ > b₂ > s₁
D₁ > pow₂ > ln₁ > div₂ > s₁
D₁ > pow₂ > 1 > s₁
D₁ > 2 > s₁
t > h > s₁
constant > s₁

Status:

D₁: [1]
t: multiset
s₁: multiset
h: multiset
constant: multiset
b₂: [1,2]
c₂: [1,2]
m₂: [1,2]
opp₁: multiset
div₂: [1,2]
pow₂: [1,2]
2: multiset
ln₁: multiset
1: multiset

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

D(t) → s(h)
D(constant) → h
D(b(x, y)) → b(D(x), D(y))
D(c(x, y)) → b(c(y, D(x)), c(x, D(y)))
D(m(x, y)) → m(D(x), D(y))
D(opp(x)) → opp(D(x))
D(div(x, y)) → m(div(D(x), y), div(c(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → b(c(c(y, pow(x, m(y, 1))), D(x)), c(c(pow(x, y), ln(x)), D(y)))
b(h, x) → x
b(x, h) → x
b(s(x), s(y)) → s(s(b(x, y)))
b(b(x, y), z) → b(x, b(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