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].
Quasi-Precedence:

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

Status:

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