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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


: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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
:12(x1, x2)  =  :11(x1)
:2(x1, x2)  =  :2(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
f1(x1)  =  f1(x1)
g2(x1, x2)  =  g
a  =  a

Lexicographic Path Order [19].
Precedence:
f1 > :2 > :^11 > g
f1 > :2 > +2 > g
f1 > :2 > a > g

The following usable rules [14] were oriented:

: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))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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.