Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

F(n__a, n__b, X) → F(X, X, X)
C → A
C → B
ACTIVATE(n__a) → A
ACTIVATE(n__b) → B

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(4) Obligation:

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

F(n__a, n__b, X) → F(X, X, X)

The TRS R consists of the following rules:

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

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

(5) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:

• Matcher: [ ]
• Semiunifier: [X / c]

---

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

F(n__a, n__b, c) → F(c, c, c)
with rule F(n__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.

(6) NO