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

(0) Obligation:

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

f(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
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(X) → IF(X, c, n__f(true))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__f(X)) → F(X)

The TRS R consists of the following rules:

f(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(false, X, Y) → ACTIVATE(Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVATE(x₁)) = x₁   
POL(F(x₁)) = x₁   
POL(IF(x₁, x₂, x₃)) = x₁ + x₃   
POL(c) = 0   
POL(false) = 1   
POL(n__f(x₁)) = x₁   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

F(X) → IF(X, c, n__f(true))
ACTIVATE(n__f(X)) → F(X)

The TRS R consists of the following rules:

f(X) → if(X, c, n__f(true))
if(true, X, Y) → X
if(false, X, Y) → activate(Y)
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(6) TRUE