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) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

D¹(b(x, y)) → B(D(x), D(y))
D¹(b(x, y)) → D¹(x)
D¹(b(x, y)) → D¹(y)
D¹(c(x, y)) → B(c(y, D(x)), c(x, D(y)))
D¹(c(x, y)) → D¹(x)
D¹(c(x, y)) → D¹(y)
D¹(m(x, y)) → D¹(x)
D¹(m(x, y)) → D¹(y)
D¹(opp(x)) → D¹(x)
D¹(div(x, y)) → D¹(x)
D¹(div(x, y)) → D¹(y)
D¹(ln(x)) → D¹(x)
D¹(pow(x, y)) → B(c(c(y, pow(x, m(y, 1))), D(x)), c(c(pow(x, y), ln(x)), D(y)))
D¹(pow(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(y)
B(s(x), s(y)) → B(x, y)
B(b(x, y), z) → B(x, b(y, z))
B(b(x, y), z) → B(y, z)

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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

B(b(x, y), z) → B(x, b(y, z))
B(s(x), s(y)) → B(x, y)
B(b(x, y), z) → B(y, z)

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.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

B(b(x, y), z) → B(x, b(y, z))
B(b(x, y), z) → B(y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
B(x1, x2)  =  B(x1)
b(x1, x2)  =  b(x1, x2)
s(x1)  =  x1
D(x1)  =  D(x1)
t  =  t
h  =  h
constant  =  constant
c(x1, x2)  =  c
m(x1, x2)  =  x1
opp(x1)  =  opp(x1)
div(x1, x2)  =  div
pow(x1, x2)  =  pow
2  =  2
ln(x1)  =  x1
1  =  1

Lexicographic path order with status [LPO].
Precedence:

B₁ > b₂ > div
D₁ > b₂ > div
D₁ > h > div
D₁ > opp₁ > div
D₁ > pow > c > div
D₁ > 2 > div
D₁ > 1 > div
t > div
constant > div

Status:

c: []
t: []
b₂: [1,2]
B₁: [1]
D₁: [1]
2: []
pow: []
h: []
constant: []
opp₁: [1]
div: []
1: []

The following usable rules [FROCOS05] were oriented:

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))

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

B(s(x), s(y)) → B(x, y)

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.
We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

B(s(x), s(y)) → B(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
B(x1, x2)  =  B(x2)
s(x1)  =  s(x1)
D(x1)  =  D(x1)
t  =  t
h  =  h
constant  =  constant
b(x1, x2)  =  b(x1, x2)
c(x1, x2)  =  c
m(x1, x2)  =  m(x1, x2)
opp(x1)  =  x1
div(x1, x2)  =  div
pow(x1, x2)  =  pow
2  =  2
ln(x1)  =  ln(x1)
1  =  1

Lexicographic path order with status [LPO].
Precedence:

ln₁ > div > pow > D₁ > h
ln₁ > div > pow > D₁ > b₂ > s₁ > B₁
ln₁ > div > pow > D₁ > m₂
ln₁ > div > pow > D₁ > 2
ln₁ > div > pow > c > b₂ > s₁ > B₁

Status:

ln₁: [1]
c: []
t: []
m₂: [1,2]
b₂: [1,2]
D₁: [1]
B₁: [1]
2: []
pow: []
h: []
constant: []
div: []
s₁: [1]
1: []

The following usable rules [FROCOS05] were oriented:

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))

(9) Obligation:

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

D¹(b(x, y)) → D¹(y)
D¹(b(x, y)) → D¹(x)
D¹(c(x, y)) → D¹(x)
D¹(c(x, y)) → D¹(y)
D¹(m(x, y)) → D¹(x)
D¹(m(x, y)) → D¹(y)
D¹(opp(x)) → D¹(x)
D¹(div(x, y)) → D¹(x)
D¹(div(x, y)) → D¹(y)
D¹(ln(x)) → D¹(x)
D¹(pow(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(y)

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.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

D¹(b(x, y)) → D¹(y)
D¹(b(x, y)) → D¹(x)
D¹(c(x, y)) → D¹(x)
D¹(c(x, y)) → D¹(y)
D¹(m(x, y)) → D¹(x)
D¹(m(x, y)) → D¹(y)
D¹(div(x, y)) → D¹(x)
D¹(div(x, y)) → D¹(y)
D¹(ln(x)) → D¹(x)
D¹(pow(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
D¹(x1)  =  D¹(x1)
b(x1, x2)  =  b(x1, x2)
c(x1, x2)  =  c(x1, x2)
m(x1, x2)  =  m(x1, x2)
opp(x1)  =  x1
div(x1, x2)  =  div(x1, x2)
ln(x1)  =  ln(x1)
pow(x1, x2)  =  pow(x1, x2)
D(x1)  =  D(x1)
t  =  t
s(x1)  =  x1
h  =  h
constant  =  constant
2  =  2
1  =  1

Lexicographic path order with status [LPO].
Precedence:

D₁ > div₂ > c₂ > b₂
D₁ > div₂ > pow₂ > b₂
D₁ > div₂ > pow₂ > m₂
D₁ > div₂ > pow₂ > 1
D₁ > div₂ > 2
D₁ > ln₁
D₁ > h

Status:

ln₁: [1]
t: []
m₂: [1,2]
c₂: [1,2]
b₂: [1,2]
D₁: [1]
2: []
pow₂: [2,1]
h: []
constant: []
div₂: [1,2]
D^1₁: [1]
1: []

The following usable rules [FROCOS05] were oriented:

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))

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

D¹(opp(x)) → D¹(x)

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.
We have to consider all minimal (P,Q,R)-chains.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

D¹(opp(x)) → D¹(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
D¹(x1)  =  D¹(x1)
opp(x1)  =  opp(x1)
D(x1)  =  D(x1)
t  =  t
s(x1)  =  s(x1)
h  =  h
constant  =  constant
b(x1, x2)  =  b(x1, x2)
c(x1, x2)  =  c
m(x1, x2)  =  x2
div(x1, x2)  =  x2
pow(x1, x2)  =  pow(x1, x2)
2  =  2
ln(x1)  =  x1
1  =  1

Lexicographic path order with status [LPO].
Precedence:

D^1₁ > 2
D₁ > opp₁ > 2
D₁ > h > 2
D₁ > b₂ > s₁ > 2
D₁ > pow₂ > c > 2
t > 2
constant > 2
1 > 2

Status:

c: []
t: []
b₂: [1,2]
D₁: [1]
2: []
pow₂: [1,2]
h: []
constant: []
opp₁: [1]
D^1₁: [1]
s₁: [1]
1: []

The following usable rules [FROCOS05] were oriented:

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))

(16) Obligation:

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.

(17) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(18) TRUE