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, r3, r4, r5, r6, r7

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            g#_A(0,0) = 1
            g#_A(0,1) = 0
            g#_A(1,0) = 1
            g#_A(1,1) = 0
            c_A(0) = 0
            c_A(1) = 1
            s_A(0) = 1
            s_A(1) = 1
            f_A(0) = 0
            f_A(1) = 0
            |0|_A = 1
            true_A = 0
            |1|_A = 1
            false_A = 0
            if_A(0,0,0) = 0
            if_A(0,0,1) = 0
            if_A(0,1,0) = 0
            if_A(0,1,1) = 1
            if_A(1,0,0) = 0
            if_A(1,0,1) = 0
            if_A(1,1,0) = 0
            if_A(1,1,1) = 1
            g_A(0,0) = 1
            g_A(0,1) = 1
            g_A(1,0) = 1
            g_A(1,1) = 1
            g#!_A(0,0) = 1
            g#!_A(0,1) = 0
            g#!_A(1,0) = 0
            g#!_A(1,1) = 0
            c!_A(0) = 1
            c!_A(1) = 0
            s!_A(0) = 0
            s!_A(1) = 0
            f!_A(0) = 0
            f!_A(1) = 0
            |0|!_A = 1
            true!_A = 0
            |1|!_A = 0
            false!_A = 0
            if!_A(0,0,0) = 1
            if!_A(0,0,1) = 0
            if!_A(0,1,0) = 1
            if!_A(0,1,1) = 0
            if!_A(1,0,0) = 0
            if!_A(1,0,1) = 0
            if!_A(1,1,0) = 0
            if!_A(1,1,1) = 0
            g!_A(0,0) = 1
            g!_A(0,1) = 0
            g!_A(1,0) = 0
            g!_A(1,1) = 0
  
    partial status:
  
      pi(g#) = [2]
      pi(c) = [1]
      pi(s) = []
      pi(f) = []
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []
      pi(if) = []
      pi(g) = [2]

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

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

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            f#_A(0) = 0
            f#_A(1) = 0
            s_A(0) = 1
            s_A(1) = 1
            f_A(0) = 0
            f_A(1) = 0
            |0|_A = 0
            true_A = 0
            |1|_A = 1
            false_A = 0
            if_A(0,0,0) = 0
            if_A(0,0,1) = 0
            if_A(0,1,0) = 0
            if_A(0,1,1) = 1
            if_A(1,0,0) = 0
            if_A(1,0,1) = 0
            if_A(1,1,0) = 0
            if_A(1,1,1) = 0
            g_A(0,0) = 0
            g_A(0,1) = 1
            g_A(1,0) = 0
            g_A(1,1) = 1
            c_A(0) = 0
            c_A(1) = 0
            f#!_A(0) = 0
            f#!_A(1) = 0
            s!_A(0) = 0
            s!_A(1) = 0
            f!_A(0) = 0
            f!_A(1) = 0
            |0|!_A = 0
            true!_A = 0
            |1|!_A = 0
            false!_A = 0
            if!_A(0,0,0) = 1
            if!_A(0,0,1) = 0
            if!_A(0,1,0) = 0
            if!_A(0,1,1) = 1
            if!_A(1,0,0) = 0
            if!_A(1,0,1) = 0
            if!_A(1,1,0) = 0
            if!_A(1,1,1) = 0
            g!_A(0,0) = 1
            g!_A(0,1) = 0
            g!_A(1,0) = 1
            g!_A(1,1) = 1
            c!_A(0) = 0
            c!_A(1) = 1
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []
      pi(if) = [2, 3]
      pi(g) = []
      pi(c) = []

The following rules are removed:

  p1

Then no dependency pair remains.