Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))

Q is empty.

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

:12(z, +2(x, f1(y))) -> :12(g2(z, y), +2(x, a))
:12(:2(x, y), z) -> :12(y, z)
:12(:2(x, y), z) -> :12(x, :2(y, z))
:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)

The TRS R consists of the following rules:

:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))

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(z, +2(x, f1(y))) -> :12(g2(z, y), +2(x, a))
:12(:2(x, y), z) -> :12(y, z)
:12(:2(x, y), z) -> :12(x, :2(y, z))
:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)

The TRS R consists of the following rules:

:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

:12(:2(x, y), z) -> :12(x, :2(y, z))
:12(:2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)

The TRS R consists of the following rules:

:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))

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(:2(x, y), z) -> :12(x, :2(y, z))
:12(:2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)
Used argument filtering: :12(x1, x2)  =  x1
:2(x1, x2)  =  :2(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
f1(x1)  =  f
g2(x1, x2)  =  g
a  =  a
Used ordering: Quasi Precedence: :_2 > +_2 > g f > a > g


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
QDP
              ↳ PisEmptyProof

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

:2(:2(x, y), z) -> :2(x, :2(y, z))
:2(+2(x, y), z) -> +2(:2(x, z), :2(y, z))
:2(z, +2(x, f1(y))) -> :2(g2(z, y), +2(x, a))

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.