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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))

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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))

Q is empty.

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

GCD2(s1(x), s1(y)) -> -12(s1(max2(x, y)), s1(min2(x, y)))
GCD2(s1(x), s1(y)) -> GCD2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
GCD2(s1(x), s1(y)) -> MAX2(x, y)
GCD2(s1(x), s1(y)) -> MIN2(x, y)
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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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:

GCD2(s1(x), s1(y)) -> -12(s1(max2(x, y)), s1(min2(x, y)))
GCD2(s1(x), s1(y)) -> GCD2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))
MIN2(s1(x), s1(y)) -> MIN2(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
GCD2(s1(x), s1(y)) -> MAX2(x, y)
GCD2(s1(x), s1(y)) -> MIN2(x, y)
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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ 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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

-12(s1(x), s1(y)) -> -12(x, y)
Used argument filtering: -12(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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
            ↳ QDPAfsSolverProof
          ↳ 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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MAX2(s1(x), s1(y)) -> MAX2(x, y)
Used argument filtering: MAX2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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
            ↳ QDPAfsSolverProof
          ↳ 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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MIN2(s1(x), s1(y)) -> MIN2(x, y)
Used argument filtering: MIN2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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

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

GCD2(s1(x), s1(y)) -> GCD2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(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)
gcd2(s1(x), s1(y)) -> gcd2(-2(s1(max2(x, y)), s1(min2(x, y))), s1(min2(x, y)))

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