Termination w.r.t. Q proof of TRS/secret06tpa09.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:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

*¹₂(x, s₁(y)) -> +¹₂(x, *₂(x, y))
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
*¹₂(x, s₁(y)) -> *¹₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
+¹₂(s₁(x), y) -> +¹₂(x, y)
F₁(s₁(x)) -> MAX₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0)))))
F₁(s₁(x)) -> +¹₂(s₁(x), s₁(s₁(s₁(s₁(0)))))
F₁(s₁(x)) -> -¹₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0)))))))
F₁(s₁(x)) -> MAX₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))
F₁(s₁(x)) -> +¹₂(s₁(x), s₁(s₁(s₁(0))))
F₁(s₁(x)) -> F₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))
MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)
F₁(s₁(x)) -> *¹₂(s₁(x), s₁(x))

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(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:

*¹₂(x, s₁(y)) -> +¹₂(x, *₂(x, y))
MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)
*¹₂(x, s₁(y)) -> *¹₂(x, y)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
+¹₂(s₁(x), y) -> +¹₂(x, y)
F₁(s₁(x)) -> MAX₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0)))))
F₁(s₁(x)) -> +¹₂(s₁(x), s₁(s₁(s₁(s₁(0)))))
F₁(s₁(x)) -> -¹₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0)))))))
F₁(s₁(x)) -> MAX₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))
F₁(s₁(x)) -> +¹₂(s₁(x), s₁(s₁(s₁(0))))
F₁(s₁(x)) -> F₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))
MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)
F₁(s₁(x)) -> *¹₂(s₁(x), s₁(x))

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-¹₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(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

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

+¹₂(s₁(x), y) -> +¹₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

+¹₂(s₁(x), y) -> +¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+¹₂(x₁, x₂)) = 3·x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

*¹₂(x, s₁(y)) -> *¹₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

*¹₂(x, s₁(y)) -> *¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*¹₂(x₁, x₂)) = 3·x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MAX₂(s₁(x), s₁(y)) -> MAX₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MAX₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(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

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₁(s₁(x)) -> F₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)

The TRS R consists of the following rules:

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(0))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MIN₂(s₁(x), s₁(y)) -> MIN₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MIN₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

min₂(0, y) -> 0
min₂(x, 0) -> 0
min₂(s₁(x), s₁(y)) -> s₁(min₂(x, y))
max₂(0, y) -> y
max₂(x, 0) -> x
max₂(s₁(x), s₁(y)) -> s₁(max₂(x, y))
+₂(0, y) -> y
+₂(s₁(x), y) -> s₁(+₂(x, y))
-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
*₂(x, 0) -> 0
*₂(x, s₁(y)) -> +₂(x, *₂(x, y))
f₁(s₁(x)) -> f₁(-₂(max₂(*₂(s₁(x), s₁(x)), +₂(s₁(x), s₁(s₁(s₁(0))))), max₂(s₁(*₂(s₁(x), s₁(x))), +₂(s₁(x), s₁(s₁(s₁(s₁(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.