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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          ackin#_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (5,7)
          s_A(x1) = ((1,0),(1,1)) x1 + (7,13)
          u21#_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (6,1)
          ackin_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (4,18)
          ackout_A(x1) = (1,6)
          u21_A(x1,x2) = ((1,0),(1,1)) x1 + (2,10)
          u22_A(x1) = ((0,0),(1,0)) x1 + (2,1)
  
    precedence:
    
      ackin# = s = u21# = ackin = u21 = u22 > ackout
    
    partial status:
  
      pi(ackin#) = [1]
      pi(s) = []
      pi(u21#) = [2]
      pi(ackin) = []
      pi(ackout) = []
      pi(u21) = [1]
      pi(u22) = []

The next rules are strictly ordered:

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

  {p2}


-- 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          ackin#_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (1,1)
          s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      ackin# = s
    
    partial status:
  
      pi(ackin#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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