YES

We show the termination of the TRS R:

  average(s(x),y) -> average(x,s(y))
  average(x,s(s(s(y)))) -> s(average(s(x),y))
  average(|0|(),|0|()) -> |0|()
  average(|0|(),s(|0|())) -> |0|()
  average(|0|(),s(s(|0|()))) -> s(|0|())

-- SCC decomposition.

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

p1: average#(s(x),y) -> average#(x,s(y))
p2: average#(x,s(s(s(y)))) -> average#(s(x),y)

and R consists of:

r1: average(s(x),y) -> average(x,s(y))
r2: average(x,s(s(s(y)))) -> s(average(s(x),y))
r3: average(|0|(),|0|()) -> |0|()
r4: average(|0|(),s(|0|())) -> |0|()
r5: average(|0|(),s(s(|0|()))) -> s(|0|())

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: average#(s(x),y) -> average#(x,s(y))
p2: average#(x,s(s(s(y)))) -> average#(s(x),y)

and R consists of:

r1: average(s(x),y) -> average(x,s(y))
r2: average(x,s(s(s(y)))) -> s(average(s(x),y))
r3: average(|0|(),|0|()) -> |0|()
r4: average(|0|(),s(|0|())) -> |0|()
r5: average(|0|(),s(s(|0|()))) -> s(|0|())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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