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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{6, x1 + 3}
          if#_A(x1,x2,x3) = max{x1 + 3, x2 + 1, x3 + 1}
          c_A = 1
          n__f_A(x1) = x1 + 4
          true_A = 1
          false_A = 4
          activate#_A(x1) = max{7, x1 + 1}
    
      precedence:
      
        c > true = false > f# = n__f > if# > activate#
      
      partial status:
    
        pi(f#) = [1]
        pi(if#) = [2, 3]
        pi(c) = []
        pi(n__f) = [1]
        pi(true) = []
        pi(false) = []
        pi(activate#) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 5
          if#_A(x1,x2,x3) = max{1, x2 - 2, x3}
          c_A = 4
          n__f_A(x1) = max{1, x1}
          true_A = 4
          false_A = 0
          activate#_A(x1) = x1 + 5
    
      precedence:
      
        f# = if# = c = n__f = true = false = activate#
      
      partial status:
    
        pi(f#) = [1]
        pi(if#) = [3]
        pi(c) = []
        pi(n__f) = [1]
        pi(true) = []
        pi(false) = []
        pi(activate#) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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