YES

We show the termination of the TRS R:

  g(x,y) -> x
  g(x,y) -> y
  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: g(x,y) -> x
r2: g(x,y) -> y
r3: f(|0|(),|1|(),x) -> f(s(x),x,x)
r4: 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: g(x,y) -> x
r2: g(x,y) -> y
r3: f(|0|(),|1|(),x) -> f(s(x),x,x)
r4: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,1,1)) x3
      |0|_A() = (0,1,1,1)
      |1|_A() = (1,1,1,1)
      s_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (1,0,1,1)

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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: g(x,y) -> x
r2: g(x,y) -> y
r3: f(|0|(),|1|(),x) -> f(s(x),x,x)
r4: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

The estimated dependency graph contains the following SCCs:

  (no SCCs)