Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


QTRS
  ↳ AAECC Innermost

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

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

Q is empty.

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

The TRS R 2 is

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

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(s(x0), x1)
f(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(s(x), y) → F(x, s(x))
F(x, s(y)) → F(y, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(s(x0), x1)
f(x0, s(x1))

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
QDP

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(s(x0), x1)
f(x0, s(x1))

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