YES

We show the termination of the TRS R:

  |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
  |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
  |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

-- SCC decomposition.

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

p1: |:|#(|:|(x,y),z) -> |:|#(x,|:|(y,z))
p2: |:|#(|:|(x,y),z) -> |:|#(y,z)
p3: |:|#(+(x,y),z) -> |:|#(x,z)
p4: |:|#(+(x,y),z) -> |:|#(y,z)
p5: |:|#(z,+(x,f(y))) -> |:|#(g(z,y),+(x,a()))

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: |:|#(|:|(x,y),z) -> |:|#(x,|:|(y,z))
p2: |:|#(+(x,y),z) -> |:|#(y,z)
p3: |:|#(+(x,y),z) -> |:|#(x,z)
p4: |:|#(|:|(x,y),z) -> |:|#(y,z)

and R consists of:

r1: |:|(|:|(x,y),z) -> |:|(x,|:|(y,z))
r2: |:|(+(x,y),z) -> +(|:|(x,z),|:|(y,z))
r3: |:|(z,+(x,f(y))) -> |:|(g(z,y),+(x,a()))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      g > |:| > f > a > + > |:|#
    
    argument filter:
  
      pi(|:|#) = 1
      pi(|:|) = [1, 2]
      pi(+) = [1, 2]
      pi(f) = [1]
      pi(g) = 1
      pi(a) = []

The next rules are strictly ordered:

  p1, p2, p3, p4

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