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

The next rules are strictly ordered:

  p1, p2, p3

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