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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            ackin#_A(x1,x2) = x1 + ((1,0),(1,1)) x2
            s_A(x1) = ((1,0),(1,1)) x1 + (3,0)
            u21#_A(x1,x2) = x1 + x2 + (4,1)
            ackin_A(x1,x2) = ((1,0),(0,0)) x2 + (0,1)
            ackout_A(x1) = ((1,0),(1,1)) x1 + (0,2)
            u21_A(x1,x2) = x1 + (1,0)
            u22_A(x1) = ((1,0),(1,1)) x1
    
      precedence:
      
        ackout > ackin# = u21# = ackin = u21 = u22 > s
      
      partial status:
    
        pi(ackin#) = [2]
        pi(s) = [1]
        pi(u21#) = [1, 2]
        pi(ackin) = []
        pi(ackout) = [1]
        pi(u21) = [1]
        pi(u22) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            ackin#_A(x1,x2) = (0,0)
            s_A(x1) = ((1,0),(1,1)) x1
            u21#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,1)
            ackin_A(x1,x2) = (2,5)
            ackout_A(x1) = (3,8)
            u21_A(x1,x2) = ((0,0),(1,0)) x1 + (1,1)
            u22_A(x1) = ((1,0),(1,1)) x1
    
      precedence:
      
        ackin = ackout > u21 = u22 > ackin# = s = u21#
      
      partial status:
    
        pi(ackin#) = []
        pi(s) = [1]
        pi(u21#) = [2]
        pi(ackin) = []
        pi(ackout) = []
        pi(u21) = []
        pi(u22) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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