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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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))
F2(s1(x), s1(y)) -> MIN2(s1(x), s1(y))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
*12(x, s1(y)) -> *12(x, y)
F2(s1(x), s1(y)) -> -12(min2(s1(x), s1(y)), max2(s1(x), s1(y)))
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), s1(y)) -> *12(s1(x), s1(y))
+12(s1(x), y) -> +12(x, y)
MAX2(s1(x), s1(y)) -> MAX2(x, y)
F2(s1(x), s1(y)) -> MAX2(s1(x), s1(y))
F2(s1(x), s1(y)) -> F2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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:

*12(x, s1(y)) -> +12(x, *2(x, y))
F2(s1(x), s1(y)) -> MIN2(s1(x), s1(y))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
*12(x, s1(y)) -> *12(x, y)
F2(s1(x), s1(y)) -> -12(min2(s1(x), s1(y)), max2(s1(x), s1(y)))
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), s1(y)) -> *12(s1(x), s1(y))
+12(s1(x), y) -> +12(x, y)
MAX2(s1(x), s1(y)) -> MAX2(x, y)
F2(s1(x), s1(y)) -> MAX2(s1(x), s1(y))
F2(s1(x), s1(y)) -> F2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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 5 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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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 Order [17,21] with Interpretation:

POL( -12(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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
          ↳ 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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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 Order [17,21] with Interpretation:

POL( +12(x1, x2) ) = x1


POL( s1(x1) ) = x1 + 1



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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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
            ↳ 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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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 Order [17,21] with Interpretation:

POL( *12(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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
          ↳ 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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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 Order [17,21] with Interpretation:

POL( MAX2(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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 Order [17,21] with Interpretation:

POL( MIN2(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ 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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

F2(s1(x), s1(y)) -> F2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(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))
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(min2(s1(x), s1(y)), max2(s1(x), s1(y))), *2(s1(x), s1(y)))

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