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:

  max/plus interpretations on natural numbers:
  
    f#_A(x1) = max{1, x1 - 2}
    if#_A(x1,x2,x3) = max{x1 - 3, x3 + 1}
    c_A = 1
    n__f_A(x1) = x1
    n__true_A = 0
    false_A = 6
    activate#_A(x1) = max{2, x1 + 1}
    activate_A(x1) = max{2, x1 + 1}
    if_A(x1,x2,x3) = max{x1 - 4, x2, x3 + 1}
    true_A = 2
    f_A(x1) = max{1, x1}

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- 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)) -> 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 estimated dependency graph contains the following SCCs:

  {p3}


-- Reduction pair.

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

p1: activate#(n__f(X)) -> 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

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    activate#_A(x1) = x1
    n__f_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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