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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ackin#_A(x1,x2) = max{3, x1, x2 + 1}
          s_A(x1) = x1 + 5
          u21#_A(x1,x2) = max{x1, x2 + 4}
          ackin_A(x1,x2) = max{x1, x2 + 6}
          ackout_A(x1) = x1 + 5
          u21_A(x1,x2) = max{x1, x2}
          u22_A(x1) = max{5, x1 - 1}
    
      precedence:
      
        s > ackin# > u21# = ackin > ackout = u21 = u22
      
      partial status:
    
        pi(ackin#) = [1, 2]
        pi(s) = [1]
        pi(u21#) = [1, 2]
        pi(ackin) = [1, 2]
        pi(ackout) = [1]
        pi(u21) = [1, 2]
        pi(u22) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          ackin#_A(x1,x2) = max{x1 + 5, x2 + 5}
          s_A(x1) = x1 + 3
          u21#_A(x1,x2) = max{x1, x2 + 5}
          ackin_A(x1,x2) = max{x1 + 4, x2 + 1}
          ackout_A(x1) = max{8, x1 + 5}
          u21_A(x1,x2) = max{x1 + 6, x2 + 14}
          u22_A(x1) = 15
    
      precedence:
      
        ackin# = s = u21# = ackin = ackout = u21 = u22
      
      partial status:
    
        pi(ackin#) = [2]
        pi(s) = [1]
        pi(u21#) = [1, 2]
        pi(ackin) = [1, 2]
        pi(ackout) = [1]
        pi(u21) = []
        pi(u22) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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