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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          admit#_A(x1,x2) = max{0, x2 - 4}
          ._A(x1,x2) = max{6, x1 + 1, x2 + 2}
          w_A = 1
          carry_A(x1,x2,x3) = max{x1 + 1, x2 + 1, x3 + 1}
    
      precedence:
      
        admit# > . = w = carry
      
      partial status:
    
        pi(admit#) = []
        pi(.) = [2]
        pi(w) = []
        pi(carry) = [1, 2, 3]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          admit#_A(x1,x2) = 0
          ._A(x1,x2) = max{2, x2}
          w_A = 3
          carry_A(x1,x2,x3) = max{x1 + 2, x2 - 1, x3 - 1}
    
      precedence:
      
        admit# = . = w = carry
      
      partial status:
    
        pi(admit#) = []
        pi(.) = [2]
        pi(w) = []
        pi(carry) = []
    

The next rules are strictly ordered:

  p1

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