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(s(x)))
f(x, s(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(s(x)))
f(x, s(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(s(x)))
f(x, s(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(s(x)))
f(x, s(s(y))) → f(y, x)
The set Q consists of the following terms:
f(s(x0), x1)
f(x0, s(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(s(x)))
F(x, s(s(y))) → F(y, x)
The TRS R consists of the following rules:
f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, x)
The set Q consists of the following terms:
f(s(x0), x1)
f(x0, s(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(s(x)))
F(x, s(s(y))) → F(y, x)
The TRS R consists of the following rules:
f(s(x), y) → f(x, s(s(x)))
f(x, s(s(y))) → f(y, x)
The set Q consists of the following terms:
f(s(x0), x1)
f(x0, s(s(x1)))
We have to consider all minimal (P,Q,R)-chains.