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) = max{x1 + 7, x2 + 14, x3 + 15}
        |0|_A = 6
        |1|_A = 3
        s_A(x1) = max{5, x1 + 1}
  
    base order 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,x2,x3) = 0
            |0|_A() = 1
            |1|_A() = 1
            s_A(x1) = 2
            f#!_A(x1,x2,x3) = 4
            |0|!_A() = 4
            |1|!_A() = 0
            s!_A(x1) = 4
  
    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)