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:
  
      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!):
        
          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) = 1
              f#!_A(x1,x2,x3) = 3
              |0|!_A() = 2
              |1|!_A() = 4
              s!_A(x1) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1,x2,x3) = max{0, x1 - 13, x3 - 7}
            |0|_A = 9
            |1|_A = 1
            s_A(x1) = max{6, x1 + 5}
            f#!_A(x1,x2,x3) = max{x1 + 8, x2 + 6, x3 + 13}
            |0|!_A = 4
            |1|!_A = 2
            s!_A(x1) = 1
        
  
    partial status:
  
      pi(f#) = [1]
      pi(|0|) = []
      pi(|1|) = []
      pi(s) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.