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:
  
      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 - 1, x2 - 5}
            c_A(x1) = x1 + 3
            s_A(x1) = x1
            f_A(x1) = max{18, x1 - 1}
            |0|_A = 0
            true_A = 7
            |1|_A = 0
            false_A = 7
            g#!_A(x1,x2) = max{3, x1 - 17, x2 - 15}
            c!_A(x1) = max{0, x1 - 15}
            s!_A(x1) = max{2, x1 - 16}
            f!_A(x1) = max{4, x1 - 1}
            |0|!_A = 2
            true!_A = 1
            |1|!_A = 4
            false!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g#_A(x1,x2) = max{0, x1 - 2, x2 - 13}
            c_A(x1) = max{11, x1 + 7}
            s_A(x1) = max{1, x1 - 1}
            f_A(x1) = max{13, x1 - 2}
            |0|_A = 1
            true_A = 0
            |1|_A = 1
            false_A = 0
            g#!_A(x1,x2) = max{2, x1 - 14, x2 - 8}
            c!_A(x1) = max{1, x1 - 44}
            s!_A(x1) = max{4, x1 - 13}
            f!_A(x1) = max{6, x1 - 2}
            |0|!_A = 6
            true!_A = 5
            |1|!_A = 7
            false!_A = 6
        
  
    partial status:
  
      pi(g#) = []
      pi(c) = []
      pi(s) = [1]
      pi(f) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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:
  
      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) = 0
            c_A(x1) = max{4, x1 - 4}
            f_A(x1) = max{13, x1 - 5}
            |0|_A = 4
            true_A = 0
            |1|_A = 4
            false_A = 0
            g#!_A(x1,x2) = max{1, x2 - 5}
            c!_A(x1) = max{2, x1 - 4}
            f!_A(x1) = max{5, x1 - 22}
            |0|!_A = 5
            true!_A = 4
            |1|!_A = 5
            false!_A = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g#_A(x1,x2) = 0
            c_A(x1) = max{7, x1 - 7}
            f_A(x1) = max{9, x1 - 3}
            |0|_A = 2
            true_A = 0
            |1|_A = 2
            false_A = 0
            g#!_A(x1,x2) = max{2, x1 - 1, x2 - 8}
            c!_A(x1) = max{1, x1 - 33}
            f!_A(x1) = max{4, x1 - 3}
            |0|!_A = 4
            true!_A = 3
            |1|!_A = 4
            false!_A = 3
        
  
    partial status:
  
      pi(g#) = [2]
      pi(c) = [1]
      pi(f) = []
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []

The following rules are removed:

  p1

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) = x1
            s_A(x1) = max{6, x1 + 1}
            f#!_A(x1) = max{3, x1 - 2}
            s!_A(x1) = max{2, x1 - 48}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1) = x1
            s_A(x1) = x1 + 8
            f#!_A(x1) = max{2, x1 - 18}
            s!_A(x1) = max{1, x1 - 18}
        
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.