YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

p1: f#(s(x)) -> f#(x)
p2: g#(x,c(y)) -> g#(x,y)
p3: g#(x,c(y)) -> g#(x,if(f(x),c(g(s(x),y)),c(y)))
p4: g#(x,c(y)) -> if#(f(x),c(g(s(x),y)),c(y))
p5: g#(x,c(y)) -> f#(x)
p6: g#(x,c(y)) -> g#(s(x),y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p6}
  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          g#_A(x1,x2) = x1 + x2 + 5
          c_A(x1) = x1 + 4
          s_A(x1) = x1 + 3
          f_A(x1) = 2
          |0|_A() = 2
          true_A() = 1
          |1|_A() = 2
          false_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          g#_A(x1,x2) = max{0, x1 - 2, x2 - 9}
          c_A(x1) = max{7, x1 + 4}
          s_A(x1) = max{1, x1 - 1}
          f_A(x1) = max{9, x1 - 13}
          |0|_A = 12
          true_A = 0
          |1|_A = 12
          false_A = 0
          g#!_A(x1,x2) = max{4, x1 - 2, x2 - 9}
          c!_A(x1) = max{1, x1 - 10}
          s!_A(x1) = max{3, x1 - 35}
          f!_A(x1) = max{2, x1 - 36}
          |0|!_A = 2
          true!_A = 1
          |1|!_A = 2
          false!_A = 1
  
    partial status:
  
      pi(g#) = [2]
      pi(c) = []
      pi(s) = []
      pi(f) = []
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []

The following rules are removed:

  p1, p2

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(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1) = x1
          s_A(x1) = x1 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1) = x1
          s_A(x1) = x1
          f#!_A(x1) = max{12, x1 - 13}
          s!_A(x1) = max{11, x1 - 12}
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.