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:

s1(a) -> a
s1(s1(x)) -> x
s1(f2(x, y)) -> f2(s1(y), s1(x))
s1(g2(x, y)) -> g2(s1(x), s1(y))
f2(x, a) -> x
f2(a, y) -> y
f2(g2(x, y), g2(u, v)) -> g2(f2(x, u), f2(y, v))
g2(a, a) -> a

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

s1(a) -> a
s1(s1(x)) -> x
s1(f2(x, y)) -> f2(s1(y), s1(x))
s1(g2(x, y)) -> g2(s1(x), s1(y))
f2(x, a) -> x
f2(a, y) -> y
f2(g2(x, y), g2(u, v)) -> g2(f2(x, u), f2(y, v))
g2(a, a) -> 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:

S1(g2(x, y)) -> S1(y)
S1(g2(x, y)) -> S1(x)
S1(f2(x, y)) -> S1(x)
S1(g2(x, y)) -> G2(s1(x), s1(y))
F2(g2(x, y), g2(u, v)) -> G2(f2(x, u), f2(y, v))
S1(f2(x, y)) -> F2(s1(y), s1(x))
S1(f2(x, y)) -> S1(y)
F2(g2(x, y), g2(u, v)) -> F2(y, v)
F2(g2(x, y), g2(u, v)) -> F2(x, u)

The TRS R consists of the following rules:

s1(a) -> a
s1(s1(x)) -> x
s1(f2(x, y)) -> f2(s1(y), s1(x))
s1(g2(x, y)) -> g2(s1(x), s1(y))
f2(x, a) -> x
f2(a, y) -> y
f2(g2(x, y), g2(u, v)) -> g2(f2(x, u), f2(y, v))
g2(a, a) -> 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:

S1(g2(x, y)) -> S1(y)
S1(g2(x, y)) -> S1(x)
S1(f2(x, y)) -> S1(x)
S1(g2(x, y)) -> G2(s1(x), s1(y))
F2(g2(x, y), g2(u, v)) -> G2(f2(x, u), f2(y, v))
S1(f2(x, y)) -> F2(s1(y), s1(x))
S1(f2(x, y)) -> S1(y)
F2(g2(x, y), g2(u, v)) -> F2(y, v)
F2(g2(x, y), g2(u, v)) -> F2(x, u)

The TRS R consists of the following rules:

s1(a) -> a
s1(s1(x)) -> x
s1(f2(x, y)) -> f2(s1(y), s1(x))
s1(g2(x, y)) -> g2(s1(x), s1(y))
f2(x, a) -> x
f2(a, y) -> y
f2(g2(x, y), g2(u, v)) -> g2(f2(x, u), f2(y, v))
g2(a, a) -> a

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 3 less nodes.

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

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

F2(g2(x, y), g2(u, v)) -> F2(x, u)
F2(g2(x, y), g2(u, v)) -> F2(y, v)

The TRS R consists of the following rules:

s1(a) -> a
s1(s1(x)) -> x
s1(f2(x, y)) -> f2(s1(y), s1(x))
s1(g2(x, y)) -> g2(s1(x), s1(y))
f2(x, a) -> x
f2(a, y) -> y
f2(g2(x, y), g2(u, v)) -> g2(f2(x, u), f2(y, v))
g2(a, a) -> 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.


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

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


POL( F2(x1, x2) ) = 3x2 + 2



The following usable rules [14] were oriented: none



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

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

s1(a) -> a
s1(s1(x)) -> x
s1(f2(x, y)) -> f2(s1(y), s1(x))
s1(g2(x, y)) -> g2(s1(x), s1(y))
f2(x, a) -> x
f2(a, y) -> y
f2(g2(x, y), g2(u, v)) -> g2(f2(x, u), f2(y, v))
g2(a, a) -> 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.

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

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

S1(g2(x, y)) -> S1(y)
S1(g2(x, y)) -> S1(x)
S1(f2(x, y)) -> S1(x)
S1(f2(x, y)) -> S1(y)

The TRS R consists of the following rules:

s1(a) -> a
s1(s1(x)) -> x
s1(f2(x, y)) -> f2(s1(y), s1(x))
s1(g2(x, y)) -> g2(s1(x), s1(y))
f2(x, a) -> x
f2(a, y) -> y
f2(g2(x, y), g2(u, v)) -> g2(f2(x, u), f2(y, v))
g2(a, a) -> 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.


S1(g2(x, y)) -> S1(y)
S1(g2(x, y)) -> S1(x)
S1(f2(x, y)) -> S1(x)
S1(f2(x, y)) -> S1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( S1(x1) ) = 2x1 + 2


POL( f2(x1, x2) ) = x1 + 3x2 + 1


POL( g2(x1, x2) ) = 3x1 + 3x2 + 1



The following usable rules [14] were oriented: none



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

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

s1(a) -> a
s1(s1(x)) -> x
s1(f2(x, y)) -> f2(s1(y), s1(x))
s1(g2(x, y)) -> g2(s1(x), s1(y))
f2(x, a) -> x
f2(a, y) -> y
f2(g2(x, y), g2(u, v)) -> g2(f2(x, u), f2(y, v))
g2(a, a) -> 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.