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:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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(g(x, y), g(u, v)) → G(f(x, u), f(y, v))
F(g(x, y), g(u, v)) → F(y, v)
S(g(x, y)) → S(x)
F(g(x, y), g(u, v)) → F(x, u)
S(g(x, y)) → S(y)
S(f(x, y)) → S(x)
S(f(x, y)) → F(s(y), s(x))
S(f(x, y)) → S(y)
S(g(x, y)) → G(s(x), s(y))

The TRS R consists of the following rules:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(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:

F(g(x, y), g(u, v)) → G(f(x, u), f(y, v))
F(g(x, y), g(u, v)) → F(y, v)
S(g(x, y)) → S(x)
F(g(x, y), g(u, v)) → F(x, u)
S(g(x, y)) → S(y)
S(f(x, y)) → S(x)
S(f(x, y)) → F(s(y), s(x))
S(f(x, y)) → S(y)
S(g(x, y)) → G(s(x), s(y))

The TRS R consists of the following rules:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

F(g(x, y), g(u, v)) → F(y, v)
F(g(x, y), g(u, v)) → F(x, u)

The TRS R consists of the following rules:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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

POL(g(x1, x2)) = 1 + (2)x_1 + (4)x_2   
POL(F(x1, x2)) = (4)x_1 + (3)x_2   
The value of delta used in the strict ordering is 7.
The following usable rules [17] 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:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(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:

S(g(x, y)) → S(x)
S(g(x, y)) → S(y)
S(f(x, y)) → S(x)
S(f(x, y)) → S(y)

The TRS R consists of the following rules:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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.


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

POL(g(x1, x2)) = 4 + (4)x_1 + (4)x_2   
POL(f(x1, x2)) = 2 + (4)x_1 + (4)x_2   
POL(S(x1)) = (4)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] 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:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(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.