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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, 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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, 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:

F2(s1(x), s1(y)) -> P1(twice1(min2(x, y)))
F2(s1(x), s1(y)) -> TWICE1(min2(x, y))
F2(s1(x), s1(y)) -> MIN2(s1(x), s1(y))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
F2(s1(x), s1(y)) -> MIN2(x, y)
F2(s1(x), s1(y)) -> -12(max2(s1(x), s1(y)), min2(s1(x), s1(y)))
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), s1(y)) -> F2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, y))))
MAX2(s1(x), s1(y)) -> MAX2(x, y)
F2(s1(x), s1(y)) -> MAX2(s1(x), s1(y))
TWICE1(s1(x)) -> TWICE1(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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(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:

F2(s1(x), s1(y)) -> P1(twice1(min2(x, y)))
F2(s1(x), s1(y)) -> TWICE1(min2(x, y))
F2(s1(x), s1(y)) -> MIN2(s1(x), s1(y))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
F2(s1(x), s1(y)) -> MIN2(x, y)
F2(s1(x), s1(y)) -> -12(max2(s1(x), s1(y)), min2(s1(x), s1(y)))
-12(s1(x), s1(y)) -> -12(x, y)
F2(s1(x), s1(y)) -> F2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, y))))
MAX2(s1(x), s1(y)) -> MAX2(x, y)
F2(s1(x), s1(y)) -> MAX2(s1(x), s1(y))
TWICE1(s1(x)) -> TWICE1(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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, y))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, 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) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(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
          ↳ QDP
          ↳ QDP

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

TWICE1(s1(x)) -> TWICE1(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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, 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.


TWICE1(s1(x)) -> TWICE1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( TWICE1(x1) ) = max{0, x1 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(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
            ↳ 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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, 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) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(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
          ↳ 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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, 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) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(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
          ↳ QDP
          ↳ QDP

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

F2(s1(x), s1(y)) -> F2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
p1(s1(x)) -> x
f2(s1(x), s1(y)) -> f2(-2(max2(s1(x), s1(y)), min2(s1(x), s1(y))), p1(twice1(min2(x, y))))

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