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(a, n__b, X) → f(X, X, X)
c → a
c → b
b → n__b
activate(n__b) → b
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(a, n__b, X) → f(X, X, X)
c → a
c → b
b → n__b
activate(n__b) → b
activate(X) → X

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:

C → B
ACTIVATE(n__b) → B
F(a, n__b, X) → F(X, X, X)

The TRS R consists of the following rules:

f(a, n__b, X) → f(X, X, X)
c → a
c → b
b → n__b
activate(n__b) → b
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

C → B
ACTIVATE(n__b) → B
F(a, n__b, X) → F(X, X, X)

The TRS R consists of the following rules:

f(a, n__b, X) → f(X, X, X)
c → a
c → b
b → n__b
activate(n__b) → b
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ NonTerminationProof

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

F(a, n__b, X) → F(X, X, X)

The TRS R consists of the following rules:

f(a, n__b, X) → f(X, X, X)
c → a
c → b
b → n__b
activate(n__b) → b
activate(X) → X

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(a, n__b, X) → F(X, X, X)

The TRS R consists of the following rules:

f(a, n__b, X) → f(X, X, X)
c → a
c → b
b → n__b
activate(n__b) → b
activate(X) → X


s = F(c, c, X) evaluates to t =F(X, X, X)

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



Rewriting sequence

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

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

F(c, n__b, c) → F(a, n__b, c)
with rule c → a at position [0] and matcher [ ]

F(a, n__b, c) → F(c, c, c)
with rule F(a, n__b, X) → F(X, X, X)

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.