YES

We show the termination of the TRS R:

  f(X) -> if(X,c(),n__f(n__true()))
  if(true(),X,Y) -> X
  if(false(),X,Y) -> activate(Y)
  f(X) -> n__f(X)
  true() -> n__true()
  activate(n__f(X)) -> f(activate(X))
  activate(n__true()) -> true()
  activate(X) -> X

-- SCC decomposition.

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

p1: f#(X) -> if#(X,c(),n__f(n__true()))
p2: if#(false(),X,Y) -> activate#(Y)
p3: activate#(n__f(X)) -> f#(activate(X))
p4: activate#(n__f(X)) -> activate#(X)
p5: activate#(n__true()) -> true#()

and R consists of:

r1: f(X) -> if(X,c(),n__f(n__true()))
r2: if(true(),X,Y) -> X
r3: if(false(),X,Y) -> activate(Y)
r4: f(X) -> n__f(X)
r5: true() -> n__true()
r6: activate(n__f(X)) -> f(activate(X))
r7: activate(n__true()) -> true()
r8: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: f#(X) -> if#(X,c(),n__f(n__true()))
p2: if#(false(),X,Y) -> activate#(Y)
p3: activate#(n__f(X)) -> activate#(X)
p4: activate#(n__f(X)) -> f#(activate(X))

and R consists of:

r1: f(X) -> if(X,c(),n__f(n__true()))
r2: if(true(),X,Y) -> X
r3: if(false(),X,Y) -> activate(Y)
r4: f(X) -> n__f(X)
r5: true() -> n__true()
r6: activate(n__f(X)) -> f(activate(X))
r7: activate(n__true()) -> true()
r8: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3, p4

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