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:

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

The next rules are strictly ordered:

  p1

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