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: standard order
        interpretations:
          ackin#_A(x1,x2) = ((0,1),(1,0)) x1 + x2 + (2,2)
          s_A(x1) = ((1,1),(1,1)) x1 + (3,3)
          u21#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (2,0)
          ackin_A(x1,x2) = (2,2)
          ackout_A(x1) = x1 + (1,1)
          u21_A(x1,x2) = (0,0)
          u22_A(x1) = (0,0)
  
    precedence:
    
      ackin# = u21# > s = ackin = ackout > u21 = u22
    
    partial status:
  
      pi(ackin#) = [2]
      pi(s) = [1]
      pi(u21#) = [1, 2]
      pi(ackin) = []
      pi(ackout) = [1]
      pi(u21) = []
      pi(u22) = []

The next rules are strictly ordered:

  p3

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: 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}


-- 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)

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: standard order
        interpretations:
          ackin#_A(x1,x2) = ((1,0),(1,1)) x1 + (5,0)
          s_A(x1) = ((1,1),(1,1)) x1 + (7,3)
          u21#_A(x1,x2) = ((1,1),(1,1)) x2 + (6,0)
          ackin_A(x1,x2) = (3,2)
          ackout_A(x1) = (4,2)
          u21_A(x1,x2) = (2,1)
          u22_A(x1) = (1,1)
  
    precedence:
    
      ackout > u22 > ackin# = s = u21# = ackin = u21
    
    partial status:
  
      pi(ackin#) = [1]
      pi(s) = [1]
      pi(u21#) = [2]
      pi(ackin) = []
      pi(ackout) = []
      pi(u21) = []
      pi(u22) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  (no SCCs)