Termination w.r.t. Q proof of /tmp/tmpGM2i75/otto09.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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:

IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
LT(s(x), s(y)) → LT(x, y)
IF(true, x, s(y), c) → PLUS(c, s(y))
PLUS(x, s(y)) → PLUS(x, y)
HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
HELP(x, s(y), c) → LT(c, x)
QUOT(x, s(y)) → HELP(x, s(y), 0)

The TRS R consists of the following rules:

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

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:

IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
LT(s(x), s(y)) → LT(x, y)
IF(true, x, s(y), c) → PLUS(c, s(y))
PLUS(x, s(y)) → PLUS(x, y)
HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
HELP(x, s(y), c) → LT(c, x)
QUOT(x, s(y)) → HELP(x, s(y), 0)

The TRS R consists of the following rules:

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

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)

The TRS R consists of the following rules:

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

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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PLUS(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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

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

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:

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

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)) = 5/4 + (15/4)x_1
POL(LT(x₁, x₂)) = x_1 + (13/4)x_2

The value of delta used in the strict ordering is 85/16.
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:

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

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:

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

The TRS R consists of the following rules:

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

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