YES

We show the termination of the TRS R:

  f(f(x,y,z),u,f(x,y,v)) -> f(x,y,f(z,u,v))
  f(x,y,y) -> y
  f(x,y,g(y)) -> x
  f(x,x,y) -> x
  f(g(x),x,y) -> y

-- SCC decomposition.

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

p1: f#(f(x,y,z),u,f(x,y,v)) -> f#(x,y,f(z,u,v))
p2: f#(f(x,y,z),u,f(x,y,v)) -> f#(z,u,v)

and R consists of:

r1: f(f(x,y,z),u,f(x,y,v)) -> f(x,y,f(z,u,v))
r2: f(x,y,y) -> y
r3: f(x,y,g(y)) -> x
r4: f(x,x,y) -> x
r5: f(g(x),x,y) -> y

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#(f(x,y,z),u,f(x,y,v)) -> f#(x,y,f(z,u,v))
p2: f#(f(x,y,z),u,f(x,y,v)) -> f#(z,u,v)

and R consists of:

r1: f(f(x,y,z),u,f(x,y,v)) -> f(x,y,f(z,u,v))
r2: f(x,y,y) -> y
r3: f(x,y,g(y)) -> x
r4: f(x,x,y) -> x
r5: f(g(x),x,y) -> y

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1,x2,x3) = ((0,1),(1,0)) x1 + ((0,1),(0,0)) x3
      f_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x3 + (1,1)
      g_A(x1) = ((1,1),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.