Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DOUBLE1(s1(x)) -> DOUBLE1(x)
EXP1(s1(x)) -> DOUBLE1(exp1(x))
HALF1(0) -> DOUBLE1(0)
EXP1(s1(x)) -> EXP1(x)
F3(b, y, x) -> EXP1(y)
TOWER1(x) -> F3(a, x, s1(0))
F3(b, y, x) -> F3(a, half1(x), exp1(y))
HALF1(s1(s1(x))) -> HALF1(x)
HALF1(s1(0)) -> HALF1(0)
F3(b, y, x) -> HALF1(x)
F3(a, s1(x), y) -> F3(b, y, s1(x))

The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

DOUBLE1(s1(x)) -> DOUBLE1(x)
EXP1(s1(x)) -> DOUBLE1(exp1(x))
HALF1(0) -> DOUBLE1(0)
EXP1(s1(x)) -> EXP1(x)
F3(b, y, x) -> EXP1(y)
TOWER1(x) -> F3(a, x, s1(0))
F3(b, y, x) -> F3(a, half1(x), exp1(y))
HALF1(s1(s1(x))) -> HALF1(x)
HALF1(s1(0)) -> HALF1(0)
F3(b, y, x) -> HALF1(x)
F3(a, s1(x), y) -> F3(b, y, s1(x))

The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

DOUBLE1(s1(x)) -> DOUBLE1(x)

The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


DOUBLE1(s1(x)) -> DOUBLE1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = 3x1 + 3


POL( DOUBLE1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


HALF1(s1(s1(x))) -> HALF1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( HALF1(x1) ) = max{0, 3x1 - 1}


POL( s1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

EXP1(s1(x)) -> EXP1(x)

The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


EXP1(s1(x)) -> EXP1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = 3x1 + 3


POL( EXP1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F3(b, y, x) -> F3(a, half1(x), exp1(y))
F3(a, s1(x), y) -> F3(b, y, s1(x))

The TRS R consists of the following rules:

tower1(x) -> f3(a, x, s1(0))
f3(a, 0, y) -> y
f3(a, s1(x), y) -> f3(b, y, s1(x))
f3(b, y, x) -> f3(a, half1(x), exp1(y))
exp1(0) -> s1(0)
exp1(s1(x)) -> double1(exp1(x))
double1(0) -> 0
double1(s1(x)) -> s1(s1(double1(x)))
half1(0) -> double1(0)
half1(s1(0)) -> half1(0)
half1(s1(s1(x))) -> s1(half1(x))

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