YES

We show the termination of the TRS R:

  ackin(s(X),s(Y)) -> u21(ackin(s(X),Y),X)
  u21(ackout(X),Y) -> u22(ackin(Y,X))

-- SCC decomposition.

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

p1: ackin#(s(X),s(Y)) -> u21#(ackin(s(X),Y),X)
p2: ackin#(s(X),s(Y)) -> ackin#(s(X),Y)
p3: u21#(ackout(X),Y) -> ackin#(Y,X)

and R consists of:

r1: ackin(s(X),s(Y)) -> u21(ackin(s(X),Y),X)
r2: u21(ackout(X),Y) -> u22(ackin(Y,X))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: ackin#(s(X),s(Y)) -> u21#(ackin(s(X),Y),X)
p2: u21#(ackout(X),Y) -> ackin#(Y,X)
p3: ackin#(s(X),s(Y)) -> ackin#(s(X),Y)

and R consists of:

r1: ackin(s(X),s(Y)) -> u21(ackin(s(X),Y),X)
r2: u21(ackout(X),Y) -> u22(ackin(Y,X))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      ackin#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,4)
      s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      u21#_A(x1,x2) = ((1,0),(1,1)) x1
      ackin_A(x1,x2) = ((1,0),(1,1)) x2 + (1,1)
      ackout_A(x1) = ((1,0),(1,1)) x1 + (2,1)
      u21_A(x1,x2) = ((1,0),(1,1)) x1 + (0,2)
      u22_A(x1) = ((1,0),(1,1)) x1

The next rules are strictly ordered:

  p1, p2, p3

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