Termination w.r.t. Q proof of /tmp/tmpDiIIoE/tower.trs

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:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

TOWERITER(s(x), y, z) → TOWERITER(p(s(x)), y, exp(y, z))
TIMES(s(x), y) → P(s(x))
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))
TOWERITER(s(x), y, z) → EXP(y, z)
P(s(s(x))) → P(s(x))
EXP(x, s(y)) → EXP(x, y)
TOWERITER(s(x), y, z) → P(s(x))
EXP(x, s(y)) → TIMES(x, exp(x, y))
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
TOWER(x, y) → TOWERITER(x, y, s(0))
TIMES(s(x), y) → TIMES(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

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:

TOWERITER(s(x), y, z) → TOWERITER(p(s(x)), y, exp(y, z))
TIMES(s(x), y) → P(s(x))
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))
TOWERITER(s(x), y, z) → EXP(y, z)
P(s(s(x))) → P(s(x))
EXP(x, s(y)) → EXP(x, y)
TOWERITER(s(x), y, z) → P(s(x))
EXP(x, s(y)) → TIMES(x, exp(x, y))
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
TOWER(x, y) → TOWERITER(x, y, s(0))
TIMES(s(x), y) → TIMES(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

P(s(s(x))) → P(s(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x₁)) = (2)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

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

          ↳ QDP

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

PLUS(s(x), y) → PLUS(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TIMES(s(x), y) → TIMES(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

EXP(x, s(y)) → EXP(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

EXP(x, s(y)) → EXP(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(EXP(x₁, x₂)) = (2)x_2
POL(s(x₁)) = 1/4 + (7/2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

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

          ↳ QDP

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

TOWERITER(s(x), y, z) → TOWERITER(p(s(x)), y, exp(y, z))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
tower(x, y) → towerIter(x, y, s(0))
towerIter(0, y, z) → z
towerIter(s(x), y, z) → towerIter(p(s(x)), y, exp(y, z))

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