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:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)


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

F2(c, c) -> F2(a, a)
F2(s1(X), c) -> F2(X, c)
F2(a, b) -> F2(s1(a), c)
F2(a, a) -> F2(a, b)

The TRS R consists of the following rules:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ QDPOrderProof

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

F2(c, c) -> F2(a, a)
F2(s1(X), c) -> F2(X, c)
F2(a, b) -> F2(s1(a), c)
F2(a, a) -> F2(a, b)

The TRS R consists of the following rules:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F2(c, c) -> F2(a, a)
The remaining pairs can at least by weakly be oriented.

F2(s1(X), c) -> F2(X, c)
F2(a, b) -> F2(s1(a), c)
F2(a, a) -> F2(a, b)
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  x1
c  =  c
a  =  a
s1(x1)  =  x1
b  =  b

Lexicographic Path Order [19].
Precedence:
c > [a, b]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ DependencyGraphProof

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

F2(s1(X), c) -> F2(X, c)
F2(a, b) -> F2(s1(a), c)
F2(a, a) -> F2(a, b)

The TRS R consists of the following rules:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
QDP
                  ↳ QDPOrderProof

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

F2(s1(X), c) -> F2(X, c)

The TRS R consists of the following rules:

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F2(s1(X), c) -> F2(X, c)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F1(x1)
s1(x1)  =  s1(x1)
c  =  c

Lexicographic Path Order [19].
Precedence:
[s1, c] > F1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
QDP
                      ↳ PisEmptyProof

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

f2(a, a) -> f2(a, b)
f2(a, b) -> f2(s1(a), c)
f2(s1(X), c) -> f2(X, c)
f2(c, c) -> f2(a, a)

The set Q consists of the following terms:

f2(a, a)
f2(a, b)
f2(s1(x0), c)
f2(c, c)

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