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:

f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> y

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> 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:

F3(s1(a), s1(b), x) -> F3(x, x, x)
G1(f3(s1(x), s1(y), z)) -> F3(x, y, z)
G1(f3(s1(x), s1(y), z)) -> G1(f3(x, y, z))

The TRS R consists of the following rules:

f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> 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:

F3(s1(a), s1(b), x) -> F3(x, x, x)
G1(f3(s1(x), s1(y), z)) -> F3(x, y, z)
G1(f3(s1(x), s1(y), z)) -> G1(f3(x, y, z))

The TRS R consists of the following rules:

f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
          ↳ QDP

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

F3(s1(a), s1(b), x) -> F3(x, x, x)

The TRS R consists of the following rules:

f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> y

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP

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

G1(f3(s1(x), s1(y), z)) -> G1(f3(x, y, z))

The TRS R consists of the following rules:

f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> y

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