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(0, 1, x) → f(x, x, x)
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, x) → f(x, x, x)
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(0, 1, x) → f(x, x, x)
The signature Sigma is {f}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, x) → f(x, x, x)
The set Q consists of the following terms:
f(0, 1, x0)
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(0, 1, x) → F(x, x, x)
The TRS R consists of the following rules:
f(0, 1, x) → f(x, x, x)
The set Q consists of the following terms:
f(0, 1, x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(x, x, x)
The TRS R consists of the following rules:
f(0, 1, x) → f(x, x, x)
The set Q consists of the following terms:
f(0, 1, x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.