YES
We show the termination of the TRS R:

  f(|0|(),|1|(),x) -> f(s(x),x,x)
  f(x,y,s(z)) -> s(f(|0|(),|1|(),z))


-- SCC decomposition.

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

p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x)
p2: f#(x,y,s(z)) -> f#(|0|(),|1|(),z)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x)
p2: f#(x,y,s(z)) -> f#(|0|(),|1|(),z)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1,x2,x3) = 0
        |0|_A = 0
        |1|_A = 9
        s_A(x1) = max{0, x1 - 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2,x3) = max{0, x1 - 20, x2 - 13, x3 - 13}
          |0|_A = 19
          |1|_A = 12
          s_A(x1) = max{11, x1 + 7}
          f#!_A(x1,x2,x3) = max{2, x1 - 21, x3 - 8}
          |0|!_A = 1
          |1|!_A = 4
          s!_A(x1) = max{3, x1 - 7}
  
    partial status:
  
      pi(f#) = []
      pi(|0|) = []
      pi(|1|) = []
      pi(s) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

The estimated dependency graph contains the following SCCs:

  (no SCCs)