YES

We show the termination of the TRS R:

  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

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(X) -> if#(X,c(),n__f(true()))
p2: if#(false(),X,Y) -> activate#(Y)
p3: activate#(n__f(X)) -> f#(X)

and R consists of:

r1: f(X) -> if(X,c(),n__f(true()))
r2: if(true(),X,Y) -> X
r3: if(false(),X,Y) -> activate(Y)
r4: f(X) -> n__f(X)
r5: activate(n__f(X)) -> f(X)
r6: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(X) -> if#(X,c(),n__f(true()))
p2: if#(false(),X,Y) -> activate#(Y)
p3: activate#(n__f(X)) -> f#(X)

and R consists of:

r1: f(X) -> if(X,c(),n__f(true()))
r2: if(true(),X,Y) -> X
r3: if(false(),X,Y) -> activate(Y)
r4: f(X) -> n__f(X)
r5: activate(n__f(X)) -> f(X)
r6: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1) = x1 + (3,0,6)
      if#_A(x1,x2,x3) = ((1,0,0),(1,0,0),(0,1,0)) x1 + ((0,0,0),(1,0,0),(0,0,0)) x2 + x3
      c_A() = (1,1,1)
      n__f_A(x1) = ((1,0,0),(1,0,0),(1,1,0)) x1 + (1,1,1)
      true_A() = (1,5,1)
      false_A() = (4,1,1)
      activate#_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (3,5,2)

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  Then no dependency pair remains.