Termination w.r.t. Q proof of /tmp/tmp7TsJex/otto11.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:

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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:

LT(s(x), s(y)) → LT(x, y)
TIMES(x, y) → HELP(x, y, 0)
HELP(x, y, c) → LT(c, y)
HELP(x, y, c) → IF(lt(c, y), x, y, c)
IF(true, x, y, c) → HELP(x, y, s(c))
PLUS(x, s(y)) → PLUS(x, y)
IF(true, x, y, c) → PLUS(x, help(x, y, s(c)))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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:

LT(s(x), s(y)) → LT(x, y)
TIMES(x, y) → HELP(x, y, 0)
HELP(x, y, c) → LT(c, y)
HELP(x, y, c) → IF(lt(c, y), x, y, c)
IF(true, x, y, c) → HELP(x, y, s(c))
PLUS(x, s(y)) → PLUS(x, y)
IF(true, x, y, c) → PLUS(x, help(x, y, s(c)))
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PLUS(x, s(y)) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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(x, s(y)) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.

PLUS(s(x), y) → PLUS(x, y)

Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(x₁, x₂)) = x_2
POL(s(x₁)) = 4 + (5/4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ 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:

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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)) = (2)x_1
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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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.

LT(s(x), s(y)) → LT(x, y)

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

POL(s(x₁)) = 1/2 + (13/4)x_1
POL(LT(x₁, x₂)) = (15/4)x_2

The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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

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

HELP(x, y, c) → IF(lt(c, y), x, y, c)
IF(true, x, y, c) → HELP(x, y, s(c))

The TRS R consists of the following rules:

times(x, y) → help(x, y, 0)
help(x, y, c) → if(lt(c, y), x, y, c)
if(true, x, y, c) → plus(x, help(x, y, s(c)))
if(false, x, y, c) → 0
lt(0, s(x)) → true
lt(s(x), 0) → false
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(0, x) → x
plus(x, s(y)) → s(plus(x, y))
plus(s(x), y) → s(plus(x, y))

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