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(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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:

MIN2(s1(x), s1(y)) -> MIN2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
GCD3(x, s1(y), s1(z)) -> GCD3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
GCD3(s1(x), s1(y), z) -> -12(max2(x, y), min2(x, y))
GCD3(s1(x), y, s1(z)) -> GCD3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
GCD3(x, s1(y), s1(z)) -> -12(max2(y, z), min2(y, z))
GCD3(s1(x), s1(y), z) -> GCD3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
GCD3(x, s1(y), s1(z)) -> MIN2(y, z)
GCD3(s1(x), y, s1(z)) -> MAX2(x, z)
GCD3(x, s1(y), s1(z)) -> MAX2(y, z)
GCD3(s1(x), s1(y), z) -> MIN2(x, y)
GCD3(s1(x), s1(y), z) -> MAX2(x, y)
MAX2(s1(x), s1(y)) -> MAX2(x, y)
GCD3(s1(x), y, s1(z)) -> MIN2(x, z)
GCD3(s1(x), y, s1(z)) -> -12(max2(x, z), min2(x, z))

The TRS R consists of the following rules:

min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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:

MIN2(s1(x), s1(y)) -> MIN2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
GCD3(x, s1(y), s1(z)) -> GCD3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
GCD3(s1(x), s1(y), z) -> -12(max2(x, y), min2(x, y))
GCD3(s1(x), y, s1(z)) -> GCD3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
GCD3(x, s1(y), s1(z)) -> -12(max2(y, z), min2(y, z))
GCD3(s1(x), s1(y), z) -> GCD3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
GCD3(x, s1(y), s1(z)) -> MIN2(y, z)
GCD3(s1(x), y, s1(z)) -> MAX2(x, z)
GCD3(x, s1(y), s1(z)) -> MAX2(y, z)
GCD3(s1(x), s1(y), z) -> MIN2(x, y)
GCD3(s1(x), s1(y), z) -> MAX2(x, y)
MAX2(s1(x), s1(y)) -> MAX2(x, y)
GCD3(s1(x), y, s1(z)) -> MIN2(x, z)
GCD3(s1(x), y, s1(z)) -> -12(max2(x, z), min2(x, z))

The TRS R consists of the following rules:

min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 9 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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( s1(x1) ) = x1 + 1


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



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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

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(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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) ) = 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

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

min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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

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(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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) ) = x1


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

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

min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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

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

GCD3(x, s1(y), s1(z)) -> GCD3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
GCD3(s1(x), y, s1(z)) -> GCD3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
GCD3(s1(x), s1(y), z) -> GCD3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)

The TRS R consists of the following rules:

min2(x, 0) -> 0
min2(0, y) -> 0
min2(s1(x), s1(y)) -> s1(min2(x, y))
max2(x, 0) -> x
max2(0, y) -> y
max2(s1(x), s1(y)) -> s1(max2(x, y))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
gcd3(s1(x), s1(y), z) -> gcd3(-2(max2(x, y), min2(x, y)), s1(min2(x, y)), z)
gcd3(x, s1(y), s1(z)) -> gcd3(x, -2(max2(y, z), min2(y, z)), s1(min2(y, z)))
gcd3(s1(x), y, s1(z)) -> gcd3(-2(max2(x, z), min2(x, z)), y, s1(min2(x, z)))
gcd3(x, 0, 0) -> x
gcd3(0, y, 0) -> y
gcd3(0, 0, z) -> z

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