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 combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        f > a > |:| > g > + > |:|#
      
      argument filter:
    
        pi(|:|#) = [1]
        pi(|:|) = [1, 2]
        pi(+) = [1, 2]
        pi(f) = [1]
        pi(g) = 1
        pi(a) = []
    
    2. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        |:|#_A(x1,x2) = ((0,0,0),(1,0,0),(1,1,0)) x1
        |:|_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + x2 + (1,2,1)
        +_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + x2 + (1,1,1)
        f_A(x1) = ((1,0,0),(1,1,0),(0,0,0)) x1 + (3,3,1)
        g_A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x1 + (1,1,1)
        a_A() = (1,1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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