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

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. 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 + (1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > ackin#
      
      argument filter:
    
        pi(ackin#) = 1
        pi(s) = 1
    

The next rules are strictly ordered:

  p1

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