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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > average#
      
      argument filter:
    
        pi(average#) = [1, 2]
        pi(s) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        average#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p2
  r1, r2, r3, r4, r5

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

  (no rules)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        average# > s
      
      argument filter:
    
        pi(average#) = [1, 2]
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        average#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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