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 strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F1(x1)
g2(x1, x2)  =  g2(x1, x2)
f2(x1, x2)  =  f2(x1, x2)
0  =  0
i1(x1)  =  i
1  =  1
2  =  2

Lexicographic Path Order [19].
Precedence:
F1 > g2
f2 > g2
0 > g2
i > g2
1 > g2
2 > g2

The following usable rules [14] were oriented:

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



↳ 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.