YES

We show the termination of the TRS R:

  a(f(),a(f(),x)) -> a(x,x)
  a(h(),x) -> a(f(),a(g(),a(f(),x)))

-- SCC decomposition.

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

p1: a#(f(),a(f(),x)) -> a#(x,x)
p2: a#(h(),x) -> a#(f(),a(g(),a(f(),x)))
p3: a#(h(),x) -> a#(g(),a(f(),x))
p4: a#(h(),x) -> a#(f(),x)

and R consists of:

r1: a(f(),a(f(),x)) -> a(x,x)
r2: a(h(),x) -> a(f(),a(g(),a(f(),x)))

The estimated dependency graph contains the following SCCs:

  {p1, p4}


-- Reduction pair.

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

p1: a#(f(),a(f(),x)) -> a#(x,x)
p2: a#(h(),x) -> a#(f(),x)

and R consists of:

r1: a(f(),a(f(),x)) -> a(x,x)
r2: a(h(),x) -> a(f(),a(g(),a(f(),x)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a#_A(x1,x2) = ((1,0),(1,1)) x2 + (2,2)
          f_A() = (3,3)
          a_A(x1,x2) = ((1,1),(0,0)) x2 + (1,1)
          h_A() = (1,1)
  
    precedence:
    
      a# > f > a > h
    
    partial status:
  
      pi(a#) = []
      pi(f) = []
      pi(a) = []
      pi(h) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(h(),x) -> a#(f(),x)

and R consists of:

r1: a(f(),a(f(),x)) -> a(x,x)
r2: a(h(),x) -> a(f(),a(g(),a(f(),x)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)