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 oriented strictly and are deleted.


:12(+2(x, y), z) -> :12(y, z)
:12(+2(x, y), z) -> :12(x, z)
The remaining pairs can at least be oriented weakly.

:12(:2(x, y), z) -> :12(x, :2(y, z))
:12(:2(x, y), z) -> :12(y, z)
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + 2·x1 + x2   
POL(:2(x1, x2)) = 2·x1 + 2·x2   
POL(:12(x1, x2)) = x1   
POL(a) = 1   
POL(f1(x1)) = 2 + 2·x1   
POL(g2(x1, x2)) = 1 + 2·x1 + 2·x2   

The following usable rules [14] were oriented: none



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

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

:12(:2(x, y), z) -> :12(y, z)
:12(:2(x, y), z) -> :12(x, :2(y, 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 oriented strictly and are deleted.


:12(:2(x, y), z) -> :12(y, z)
:12(:2(x, y), z) -> :12(x, :2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = x2   
POL(:2(x1, x2)) = 2 + x1 + x2   
POL(:12(x1, x2)) = 2·x1 + x2   
POL(a) = 0   
POL(f1(x1)) = 1 + 2·x1   
POL(g2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented:

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



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