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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g#_A(x1,x2) = max{0, x1 - 29, x2 - 17}
            c_A(x1) = max{15, x1 + 4}
            s_A(x1) = max{28, x1 - 2}
            g#!_A(x1,x2) = max{5, x2 - 9}
            c!_A(x1) = x1 + 6
            s!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              g#_A(x1,x2) = x1 + x2
              c_A(x1) = x1 + 1
              s_A(x1) = x1 + 1
              g#!_A(x1,x2) = 7
              c!_A(x1) = 9
              s!_A(x1) = 6
        
  
    partial status:
  
      pi(g#) = [1]
      pi(c) = []
      pi(s) = []

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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1) = 0
            s_A(x1) = 20
            f#!_A(x1) = 0
            s!_A(x1) = max{23, x1 - 22}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1) = x1
              s_A(x1) = x1 + 2
              f#!_A(x1) = x1 + 3
              s!_A(x1) = 4
        
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.