Termination w.r.t. Q proof of /tmp/tmpZ9FStW/otto12.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(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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:

TIMES(s(x), y) → PLUS(y, times(x, y))
HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
TOWER(x, y) → TOWERITER(0, x, y, s(0))
TIMES(s(x), y) → TIMES(x, y)
HELP(false, c, x, y, z) → EXP(y, z)
TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)
EXP(x, s(y)) → EXP(x, y)
TOWERITER(c, x, y, z) → GE(c, x)
EXP(x, s(y)) → TIMES(x, exp(x, y))
GE(s(x), s(y)) → GE(x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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:

TIMES(s(x), y) → PLUS(y, times(x, y))
HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
TOWER(x, y) → TOWERITER(0, x, y, s(0))
TIMES(s(x), y) → TIMES(x, y)
HELP(false, c, x, y, z) → EXP(y, z)
TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)
EXP(x, s(y)) → EXP(x, y)
TOWERITER(c, x, y, z) → GE(c, x)
EXP(x, s(y)) → TIMES(x, exp(x, y))
GE(s(x), s(y)) → GE(x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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.

GE(s(x), s(y)) → GE(x, y)

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

POL(GE(x₁, x₂)) = x_2
POL(s(x₁)) = 1 + 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(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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.

PLUS(s(x), y) → PLUS(x, y)

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

POL(PLUS(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ 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(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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.

TIMES(s(x), y) → TIMES(x, y)

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

POL(TIMES(x₁, x₂)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ 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(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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

            ↳ 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(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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)) = (1/4)x_2
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
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(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), 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:

HELP(false, c, x, y, z) → TOWERITER(s(c), x, y, exp(y, z))
TOWERITER(c, x, y, z) → HELP(ge(c, x), c, x, y, z)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
exp(x, 0) → s(0)
exp(x, s(y)) → times(x, exp(x, y))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
tower(x, y) → towerIter(0, x, y, s(0))
towerIter(c, x, y, z) → help(ge(c, x), c, x, y, z)
help(true, c, x, y, z) → z
help(false, c, x, y, z) → towerIter(s(c), x, y, exp(y, z))

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