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(s(x), s(y)) → B(x, y)
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, x2)
b(x1, x2)  =  b(x1, x2)
s(x1)  =  s(x1)
D(x1)  =  D(x1)
t  =  t
h  =  h
constant  =  constant
c(x1, x2)  =  c
m(x1, x2)  =  m(x1, x2)
opp(x1)  =  x1
div(x1, x2)  =  div(x1, x2)
pow(x1, x2)  =  pow
2  =  2
ln(x1)  =  ln(x1)
1  =  1

Lexicographic path order with status [LPO].
Quasi-Precedence:

B₂ > [b₂, m₂, pow] > [s₁, t] > c
B₂ > [b₂, m₂, pow] > 1 > c
constant > c
ln₁ > [D₁, div₂] > [b₂, m₂, pow] > [s₁, t] > c
ln₁ > [D₁, div₂] > [b₂, m₂, pow] > 1 > c
ln₁ > [D₁, div₂] > h > c
ln₁ > [D₁, div₂] > 2 > c

Status:

B₂: [1,2]
b₂: [1,2]
s₁: [1]
D₁: [1]
t: []
h: []
constant: []
c: []
m₂: [2,1]
div₂: [2,1]
pow: []
2: []
ln₁: [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))

(7) 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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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.

(11) 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].
Quasi-Precedence:

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

Status:

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

(12) 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.

(13) 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)  =  m(x1, x2)
div(x1, x2)  =  div(x1, x2)
pow(x1, x2)  =  pow
2  =  2
ln(x1)  =  ln(x1)
1  =  1

Lexicographic path order with status [LPO].
Quasi-Precedence:

constant > c
2 > c
ln₁ > [D₁, t, h, div₂] > opp₁ > D^1₁ > c
ln₁ > [D₁, t, h, div₂] > b₂ > s₁ > c
ln₁ > [D₁, t, h, div₂] > [m₂, pow] > c
1 > c

Status:

D^1₁: [1]
opp₁: [1]
D₁: [1]
t: []
s₁: [1]
h: []
constant: []
b₂: [1,2]
c: []
m₂: [1,2]
div₂: [1,2]
pow: []
2: []
ln₁: [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:
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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE