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:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

:1(z, +(x, f(y))) → :1(g(z, y), +(x, a))
:1(+(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(x, :(y, z))
:1(+(x, y), z) → :1(x, z)

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(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:

:1(z, +(x, f(y))) → :1(g(z, y), +(x, a))
:1(+(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(x, :(y, z))
:1(+(x, y), z) → :1(x, z)

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

:1(+(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(x, :(y, z))
:1(+(x, y), z) → :1(x, z)

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


:1(+(x, y), z) → :1(y, z)
:1(+(x, y), z) → :1(x, z)
The remaining pairs can at least be oriented weakly.

:1(:(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(x, :(y, z))
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = 0   
POL(a) = 0   
POL(:(x1, x2)) = (4)x_1 + x_2   
POL(g(x1, x2)) = 0   
POL(:1(x1, x2)) = (1/4)x_1   
POL(+(x1, x2)) = 1 + x_1 + (2)x_2   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none



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

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

:1(:(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(x, :(y, z))

The TRS R consists of the following rules:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


:1(:(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(x, :(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = 0   
POL(a) = 0   
POL(:(x1, x2)) = 1/2 + (4)x_1 + x_2   
POL(g(x1, x2)) = 0   
POL(:1(x1, x2)) = (4)x_1 + (1/4)x_2   
POL(+(x1, x2)) = x_1   
The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented:

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))



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

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

:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(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.