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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 NonTerminationProof (⇔)
↳4 NO

(0) Obligation:

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

f(g(X), Y) → f(X, n__f(g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
activate(n__f(X1, X2)) → f(X1, X2)
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(g(X), Y) → F(X, n__f(g(X), activate(Y)))
F(g(X), Y) → ACTIVATE(Y)
ACTIVATE(n__f(X1, X2)) → F(X1, X2)

The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
activate(n__f(X1, X2)) → f(X1, X2)
activate(X) → X

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

(3) 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(g(X'), n__f(g(X), X2)) evaluates to t =F(X, n__f(g(X), activate(X2)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [X' / X'', X2 / activate(X2)]
  • Semiunifier: [X / g(X'')]



Rewriting sequence

F(g(X'), n__f(g(g(X'')), X2)) → ACTIVATE(n__f(g(g(X'')), X2))
with rule F(g(X'''), Y) → ACTIVATE(Y) at position [] and matcher [X''' / X', Y / n__f(g(g(X'')), X2)]

ACTIVATE(n__f(g(g(X'')), X2)) → F(g(g(X'')), X2)
with rule ACTIVATE(n__f(X1, X2')) → F(X1, X2') at position [] and matcher [X1 / g(g(X'')), X2' / X2]

F(g(g(X'')), X2) → F(g(X''), n__f(g(g(X'')), activate(X2)))
with rule F(g(X), Y) → F(X, n__f(g(X), activate(Y)))

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.



(4) NO