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:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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:

F(f(x, y), z) → F(y, z)
F(g(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(f(x, y), z) → F(x, f(y, z))

The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

F(f(x, y), z) → F(y, z)
F(g(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(f(x, y), z) → F(x, f(y, z))

The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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.


F(g(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
The remaining pairs can at least be oriented weakly.

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

POL(i(x1)) = 0   
POL(g(x1, x2)) = 1/4 + (4)x_1 + (2)x_2   
POL(f(x1, x2)) = (4)x_1 + (2)x_2   
POL(2) = 0   
POL(F(x1, x2)) = (1/4)x_1   
POL(0) = 0   
POL(1) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



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

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

F(f(x, y), z) → F(y, z)
F(f(x, y), z) → F(x, f(y, z))

The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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.


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

POL(i(x1)) = 0   
POL(g(x1, x2)) = 0   
POL(f(x1, x2)) = 4 + (2)x_1 + (4)x_2   
POL(2) = 0   
POL(F(x1, x2)) = (4)x_1   
POL(0) = 0   
POL(1) = 0   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



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

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

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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.