Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

f(X, g(X), Y) → f(Y, Y, Y)
g(b) → c
b → c

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(X, g(X), Y) → f(Y, Y, Y)
g(b) → c
b → c

Q is empty.

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(X, g(X), Y) → F(Y, Y, Y)

The TRS R consists of the following rules:

f(X, g(X), Y) → f(Y, Y, Y)
g(b) → c
b → c

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation

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

F(X, g(X), Y) → F(Y, Y, Y)

The TRS R consists of the following rules:

f(X, g(X), Y) → f(Y, Y, Y)
g(b) → c
b → c

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(X, g(X), Y) → F(Y, Y, Y) we obtained the following new rules:

F(x0, g(x0), g(y_1)) → F(g(y_1), g(y_1), g(y_1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ NonTerminationProof

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

F(x0, g(x0), g(y_1)) → F(g(y_1), g(y_1), g(y_1))

The TRS R consists of the following rules:

f(X, g(X), Y) → f(Y, Y, Y)
g(b) → c
b → c

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

F(x0, g(x0), g(y_1)) → F(g(y_1), g(y_1), g(y_1))

The TRS R consists of the following rules:

f(X, g(X), Y) → f(Y, Y, Y)
g(b) → c
b → c


s = F(g(b), g(b), g(y_1)) evaluates to t =F(g(y_1), g(y_1), g(y_1))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

F(g(b), g(b), g(b)) → F(g(b), g(c), g(b))
with rule b → c at position [1,0] and matcher [ ]

F(g(b), g(c), g(b)) → F(c, g(c), g(b))
with rule g(b) → c at position [0] and matcher [ ]

F(c, g(c), g(b)) → F(g(b), g(b), g(b))
with rule F(x0, g(x0), g(y_1)) → F(g(y_1), g(y_1), g(y_1))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.