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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = ((1,0),(1,0)) x1 + (2,3)
            if#_A(x1,x2,x3) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x3
            c_A() = (1,0)
            n__f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            true_A() = (0,1)
            false_A() = (4,4)
            activate#_A(x1) = ((1,0),(1,0)) x1 + (2,3)
    
      precedence:
      
        f# = c = n__f = true > if# = false = activate#
      
      partial status:
    
        pi(f#) = []
        pi(if#) = [3]
        pi(c) = []
        pi(n__f) = []
        pi(true) = []
        pi(false) = []
        pi(activate#) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = (2,2)
            if#_A(x1,x2,x3) = (1,1)
            c_A() = (0,0)
            n__f_A(x1) = (3,3)
            true_A() = (1,1)
            false_A() = (4,4)
            activate#_A(x1) = (3,3)
    
      precedence:
      
        true > n__f > false > activate# > f# = c > if#
      
      partial status:
    
        pi(f#) = []
        pi(if#) = []
        pi(c) = []
        pi(n__f) = []
        pi(true) = []
        pi(false) = []
        pi(activate#) = []
    

The next rules are strictly ordered:

  p2, p3

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(true()))

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:

  (no SCCs)