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.

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 contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F2(g2(x, y), g2(u, v)) -> F2(x, u)
F2(g2(x, y), g2(u, v)) -> F2(y, v)
Used argument filtering: F2(x1, x2)  =  x2
g2(x1, x2)  =  g2(x1, x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S1(g2(x, y)) -> S1(y)
S1(g2(x, y)) -> S1(x)
S1(f2(x, y)) -> S1(x)
S1(f2(x, y)) -> S1(y)
Used argument filtering: S1(x1)  =  x1
g2(x1, x2)  =  g2(x1, x2)
f2(x1, x2)  =  f2(x1, x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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.