YES

We show the termination of the TRS R:

  f(|0|()) -> true()
  f(|1|()) -> false()
  f(s(x)) -> f(x)
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  g(s(x),s(y)) -> if(f(x),s(x),s(y))
  g(x,c(y)) -> g(x,g(s(c(y)),y))

-- SCC decomposition.

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

p1: f#(s(x)) -> f#(x)
p2: g#(s(x),s(y)) -> if#(f(x),s(x),s(y))
p3: g#(s(x),s(y)) -> f#(x)
p4: g#(x,c(y)) -> g#(x,g(s(c(y)),y))
p5: g#(x,c(y)) -> g#(s(c(y)),y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),x,y) -> x
r5: if(false(),x,y) -> y
r6: g(s(x),s(y)) -> if(f(x),s(x),s(y))
r7: g(x,c(y)) -> g(x,g(s(c(y)),y))

The estimated dependency graph contains the following SCCs:

  {p4, p5}
  {p1}


-- Reduction pair.

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

p1: g#(x,c(y)) -> g#(s(c(y)),y)
p2: g#(x,c(y)) -> g#(x,g(s(c(y)),y))

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),x,y) -> x
r5: if(false(),x,y) -> y
r6: g(s(x),s(y)) -> if(f(x),s(x),s(y))
r7: g(x,c(y)) -> g(x,g(s(c(y)),y))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = max{x1 + 2, x2 + 4}
          c_A(x1) = x1 + 8
          s_A(x1) = 0
          g_A(x1,x2) = x2 + 7
          f_A(x1) = 4
          |0|_A = 4
          true_A = 0
          |1|_A = 0
          false_A = 1
          if_A(x1,x2,x3) = max{x1 + 2, x2 + 7, x3 + 4}
    
      precedence:
      
        g# = c = true = false > g > f = |0| = |1| > if > s
      
      partial status:
    
        pi(g#) = [2]
        pi(c) = [1]
        pi(s) = []
        pi(g) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = [1, 2, 3]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1,x2) = x2 + 20
          c_A(x1) = max{18, x1 + 6}
          s_A(x1) = 22
          g_A(x1,x2) = 5
          f_A(x1) = 23
          |0|_A = 23
          true_A = 24
          |1|_A = 23
          false_A = 24
          if_A(x1,x2,x3) = max{20, x1 - 1, x2 + 1, x3 + 1}
    
      precedence:
      
        g# = c = s = g = f = true = |1| > |0| = false > if
      
      partial status:
    
        pi(g#) = []
        pi(c) = [1]
        pi(s) = []
        pi(g) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: f#(s(x)) -> f#(x)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),x,y) -> x
r5: if(false(),x,y) -> y
r6: g(s(x),s(y)) -> if(f(x),s(x),s(y))
r7: g(x,c(y)) -> g(x,g(s(c(y)),y))

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{4, x1 + 3}
          s_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        f# = s
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{1, x1 - 1}
          s_A(x1) = x1
    
      precedence:
      
        f# = s
      
      partial status:
    
        pi(f#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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