Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f1(X) -> if3(X, c, n__f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(X) -> if3(X, c, n__f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVATE1(n__f1(X)) -> F1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
F1(X) -> IF3(X, c, n__f1(true))

The TRS R consists of the following rules:

f1(X) -> if3(X, c, n__f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X

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

↳ QTRS
  ↳ DependencyPairsProof
QDP

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

ACTIVATE1(n__f1(X)) -> F1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
F1(X) -> IF3(X, c, n__f1(true))

The TRS R consists of the following rules:

f1(X) -> if3(X, c, n__f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X

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