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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        u22 > u21 > ackin > ackin# > ackout > u21# > s
      
      argument filter:
    
        pi(ackin#) = 2
        pi(s) = 1
        pi(u21#) = 1
        pi(ackin) = 2
        pi(ackout) = 1
        pi(u21) = 1
        pi(u22) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        u22 > ackin > u21 > ackin# > ackout > u21# > s
      
      argument filter:
    
        pi(ackin#) = 2
        pi(s) = 1
        pi(u21#) = 1
        pi(ackin) = 2
        pi(ackout) = [1]
        pi(u21) = 1
        pi(u22) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        u22 > ackin > u21 > ackin# > ackout > u21# > s
      
      argument filter:
    
        pi(ackin#) = [2]
        pi(s) = [1]
        pi(u21#) = []
        pi(ackin) = []
        pi(ackout) = [1]
        pi(u21) = 1
        pi(u22) = 1
    

The next rules are strictly ordered:

  p1, p2, p3

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