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:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))


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

DOUBLE(s(x)) → DOUBLE(x)
EXP(s(x)) → DOUBLE(exp(x))
HALF(0) → DOUBLE(0)
EXP(s(x)) → EXP(x)
F(b, y, x) → EXP(y)
TOWER(x) → F(a, x, s(0))
F(b, y, x) → F(a, half(x), exp(y))
HALF(s(s(x))) → HALF(x)
HALF(s(0)) → HALF(0)
F(b, y, x) → HALF(x)
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

DOUBLE(s(x)) → DOUBLE(x)
EXP(s(x)) → DOUBLE(exp(x))
HALF(0) → DOUBLE(0)
EXP(s(x)) → EXP(x)
F(b, y, x) → EXP(y)
TOWER(x) → F(a, x, s(0))
F(b, y, x) → F(a, half(x), exp(y))
HALF(s(s(x))) → HALF(x)
HALF(s(0)) → HALF(0)
F(b, y, x) → HALF(x)
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

DOUBLE(s(x)) → DOUBLE(x)
EXP(s(x)) → DOUBLE(exp(x))
HALF(0) → DOUBLE(0)
EXP(s(x)) → EXP(x)
F(b, y, x) → EXP(y)
TOWER(x) → F(a, x, s(0))
F(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))
F(b, y, x) → HALF(x)
HALF(s(0)) → HALF(0)
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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.


DOUBLE(s(x)) → DOUBLE(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DOUBLE(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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.


HALF(s(s(x))) → HALF(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

EXP(s(x)) → EXP(x)

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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.


EXP(s(x)) → EXP(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
EXP(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

F(b, y, x) → F(a, half(x), exp(y))
F(a, s(x), y) → F(b, y, s(x))

The TRS R consists of the following rules:

tower(x) → f(a, x, s(0))
f(a, 0, y) → y
f(a, s(x), y) → f(b, y, s(x))
f(b, y, x) → f(a, half(x), exp(y))
exp(0) → s(0)
exp(s(x)) → double(exp(x))
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → double(0)
half(s(0)) → half(0)
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

tower(x0)
f(a, 0, x0)
f(a, s(x0), x1)
f(b, x0, x1)
exp(0)
exp(s(x0))
double(0)
double(s(x0))
half(0)
half(s(0))
half(s(s(x0)))

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