YES

We show the termination of the TRS R:

  admit(x,nil()) -> nil()
  admit(x,.(u,.(v,.(w(),z)))) -> cond(=(sum(x,u,v),w()),.(u,.(v,.(w(),admit(carry(x,u,v),z)))))
  cond(true(),y) -> y

-- SCC decomposition.

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

p1: admit#(x,.(u,.(v,.(w(),z)))) -> cond#(=(sum(x,u,v),w()),.(u,.(v,.(w(),admit(carry(x,u,v),z)))))
p2: admit#(x,.(u,.(v,.(w(),z)))) -> admit#(carry(x,u,v),z)

and R consists of:

r1: admit(x,nil()) -> nil()
r2: admit(x,.(u,.(v,.(w(),z)))) -> cond(=(sum(x,u,v),w()),.(u,.(v,.(w(),admit(carry(x,u,v),z)))))
r3: cond(true(),y) -> y

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: admit#(x,.(u,.(v,.(w(),z)))) -> admit#(carry(x,u,v),z)

and R consists of:

r1: admit(x,nil()) -> nil()
r2: admit(x,.(u,.(v,.(w(),z)))) -> cond(=(sum(x,u,v),w()),.(u,.(v,.(w(),admit(carry(x,u,v),z)))))
r3: cond(true(),y) -> y

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: standard order
      interpretations:
        admit#_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2
        ._A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (1,0)
        w_A() = (3,0)
        carry_A(x1,x2,x3) = x1 + ((0,0),(1,0)) x2 + ((0,1),(1,1)) x3 + (1,3)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        admit#_A(x1,x2) = ((1,0),(1,0)) x2
        ._A(x1,x2) = ((1,1),(1,0)) x1 + ((1,1),(0,0)) x2 + (0,1)
        w_A() = (1,1)
        carry_A(x1,x2,x3) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        admit#_A(x1,x2) = ((0,1),(0,0)) x2
        ._A(x1,x2) = ((0,0),(1,0)) x1 + (1,0)
        w_A() = (0,1)
        carry_A(x1,x2,x3) = (1,1)
    

The next rules are strictly ordered:

  p1

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