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(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)


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(s(X), c) → F(X, c)
F(a, a) → F(a, b)
F(c, c) → F(a, a)
F(a, b) → F(s(a), c)

The TRS R consists of the following rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof

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

F(s(X), c) → F(X, c)
F(a, a) → F(a, b)
F(c, c) → F(a, a)
F(a, b) → F(s(a), c)

The TRS R consists of the following rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

F(s(X), c) → F(X, c)
F(a, a) → F(a, b)
F(c, c) → F(a, a)
F(a, b) → F(s(a), c)

R is empty.
The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPOrderProof

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

F(s(X), c) → F(X, c)
F(a, a) → F(a, b)
F(c, c) → F(a, a)
F(a, b) → F(s(a), c)

R is empty.
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(c, c) → F(a, a)
The remaining pairs can at least be oriented weakly.

F(s(X), c) → F(X, c)
F(a, a) → F(a, b)
F(a, b) → F(s(a), c)
Used ordering: Polynomial interpretation [25]:

POL(F(x1, x2)) = x1   
POL(a) = 0   
POL(b) = 0   
POL(c) = 1   
POL(s(x1)) = x1   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof

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

F(s(X), c) → F(X, c)
F(a, a) → F(a, b)
F(a, b) → F(s(a), c)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ QDPSizeChangeProof

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

F(s(X), c) → F(X, c)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: