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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) 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:

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) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• B(b(x, y), z) → B(x, b(y, z))
  The graph contains the following edges 1 > 1

• B(s(x), s(y)) → B(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

• B(b(x, y), z) → B(y, z)
  The graph contains the following edges 1 > 1, 2 >= 2

(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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(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)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• D¹(b(x, y)) → D¹(y)
  The graph contains the following edges 1 > 1

• D¹(b(x, y)) → D¹(x)
  The graph contains the following edges 1 > 1

• D¹(c(x, y)) → D¹(x)
  The graph contains the following edges 1 > 1

• D¹(c(x, y)) → D¹(y)
  The graph contains the following edges 1 > 1

• D¹(m(x, y)) → D¹(x)
  The graph contains the following edges 1 > 1

• D¹(m(x, y)) → D¹(y)
  The graph contains the following edges 1 > 1

• D¹(opp(x)) → D¹(x)
  The graph contains the following edges 1 > 1

• D¹(div(x, y)) → D¹(x)
  The graph contains the following edges 1 > 1

• D¹(div(x, y)) → D¹(y)
  The graph contains the following edges 1 > 1

• D¹(ln(x)) → D¹(x)
  The graph contains the following edges 1 > 1

• D¹(pow(x, y)) → D¹(x)
  The graph contains the following edges 1 > 1

• D¹(pow(x, y)) → D¹(y)
  The graph contains the following edges 1 > 1

(14) TRUE