Termination w.r.t. Q proof of /tmp/tmpwo7cwY/tpa06.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:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(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:

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(x), min(s(y), s(z)))
F(s(x), s(y), s(z)) → P(min(s(x), max(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(y), s(z))
F(s(x), s(y), s(z)) → MIN(s(x), max(s(y), s(z)))
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
F(x, y, 0) → MAX(x, y)
F(x, 0, z) → MAX(x, z)
F(0, y, z) → MAX(y, z)
F(s(x), s(y), s(z)) → MAX(s(y), s(z))
F(s(x), s(y), s(z)) → MAX(s(x), max(s(y), s(z)))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(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:

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(x), min(s(y), s(z)))
F(s(x), s(y), s(z)) → P(min(s(x), max(s(y), s(z))))
F(s(x), s(y), s(z)) → MIN(s(y), s(z))
F(s(x), s(y), s(z)) → MIN(s(x), max(s(y), s(z)))
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
F(x, y, 0) → MAX(x, y)
F(x, 0, z) → MAX(x, z)
F(0, y, z) → MAX(y, z)
F(s(x), s(y), s(z)) → MAX(s(y), s(z))
F(s(x), s(y), s(z)) → MAX(s(x), max(s(y), s(z)))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(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 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(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.

MAX(s(x), s(y)) → MAX(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(MAX(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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(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:

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(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.

MIN(s(x), s(y)) → MIN(x, y)

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

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

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:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(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:

F(s(x), s(y), s(z)) → F(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
p(s(x)) → x
f(s(x), s(y), s(z)) → f(max(s(x), max(s(y), s(z))), p(min(s(x), max(s(y), s(z)))), min(s(x), min(s(y), s(z))))
f(0, y, z) → max(y, z)
f(x, 0, z) → max(x, z)
f(x, y, 0) → max(x, y)

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