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:

f2(0, y) -> y
f2(x, 0) -> x
f2(i1(x), y) -> i1(x)
f2(f2(x, y), z) -> f2(x, f2(y, z))
f2(g2(x, y), z) -> g2(f2(x, z), f2(y, z))
f2(1, g2(x, y)) -> x
f2(2, g2(x, y)) -> y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f2(0, y) -> y
f2(x, 0) -> x
f2(i1(x), y) -> i1(x)
f2(f2(x, y), z) -> f2(x, f2(y, z))
f2(g2(x, y), z) -> g2(f2(x, z), f2(y, z))
f2(1, g2(x, y)) -> x
f2(2, g2(x, y)) -> y

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:

F2(g2(x, y), z) -> F2(y, z)
F2(g2(x, y), z) -> F2(x, z)
F2(f2(x, y), z) -> F2(x, f2(y, z))
F2(f2(x, y), z) -> F2(y, z)

The TRS R consists of the following rules:

f2(0, y) -> y
f2(x, 0) -> x
f2(i1(x), y) -> i1(x)
f2(f2(x, y), z) -> f2(x, f2(y, z))
f2(g2(x, y), z) -> g2(f2(x, z), f2(y, z))
f2(1, g2(x, y)) -> x
f2(2, g2(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:

F2(g2(x, y), z) -> F2(y, z)
F2(g2(x, y), z) -> F2(x, z)
F2(f2(x, y), z) -> F2(x, f2(y, z))
F2(f2(x, y), z) -> F2(y, z)

The TRS R consists of the following rules:

f2(0, y) -> y
f2(x, 0) -> x
f2(i1(x), y) -> i1(x)
f2(f2(x, y), z) -> f2(x, f2(y, z))
f2(g2(x, y), z) -> g2(f2(x, z), f2(y, z))
f2(1, g2(x, y)) -> x
f2(2, g2(x, y)) -> y

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.


F2(g2(x, y), z) -> F2(y, z)
F2(g2(x, y), z) -> F2(x, z)
F2(f2(x, y), z) -> F2(x, f2(y, z))
F2(f2(x, y), z) -> F2(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 2 ) = 0


POL( i1(x1) ) = max{0, 3x1 - 3}


POL( g2(x1, x2) ) = 2x1 + 3x2 + 3


POL( f2(x1, x2) ) = 2x1 + 2x2 + 3


POL( 1 ) = 1


POL( 0 ) = 1


POL( F2(x1, x2) ) = x1 + 2



The following usable rules [14] were oriented: none



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

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

f2(0, y) -> y
f2(x, 0) -> x
f2(i1(x), y) -> i1(x)
f2(f2(x, y), z) -> f2(x, f2(y, z))
f2(g2(x, y), z) -> g2(f2(x, z), f2(y, z))
f2(1, g2(x, y)) -> x
f2(2, g2(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.