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:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The signature Sigma is {f, g}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(c(s(x), y)) → F(c(x, s(y)))
G(c(x, s(y))) → G(c(s(x), y))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

F(c(s(x), y)) → F(c(x, s(y)))
G(c(x, s(y))) → G(c(s(x), y))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

F(c(s(x), y)) → F(c(x, s(y)))
G(c(x, s(y))) → G(c(s(x), y))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

G(c(x, s(y))) → G(c(s(x), y))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))

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.


G(c(x, s(y))) → G(c(s(x), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1)  =  x1
c(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

F(c(s(x), y)) → F(c(x, s(y)))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))

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.


F(c(s(x), y)) → F(c(x, s(y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
c(x1, x2)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

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

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))

The set Q consists of the following terms:

f(c(s(x0), x1))
g(c(x0, s(x1)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.