We consider the following Problem:

  Strict Trs:
    {  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}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  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}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {admit(x, nil()) -> nil()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(admit) = {}, Uargs(.) = {2}, Uargs(cond) = {2},
        Uargs(=) = {}, Uargs(sum) = {}, Uargs(carry) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       admit(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                       [0 0]      [0 0]      [1]
       nil() = [0]
               [0]
       .(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                   [0 0]      [0 0]      [2]
       w() = [0]
             [0]
       cond(x1, x2) = [0 0] x1 + [1 2] x2 + [0]
                      [0 0]      [0 0]      [1]
       =(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                   [0 0]      [0 0]      [0]
       sum(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                         [0 0]      [0 0]      [0 0]      [0]
       carry(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                           [0 0]      [0 0]      [0 0]      [0]
       true() = [0]
                [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  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}
      Weak Trs: {admit(x, nil()) -> nil()}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {cond(true(), y) -> y}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(admit) = {}, Uargs(.) = {2}, Uargs(cond) = {2},
          Uargs(=) = {}, Uargs(sum) = {}, Uargs(carry) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         admit(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                         [0 0]      [0 0]      [1]
         nil() = [0]
                 [0]
         .(x1, x2) = [0 0] x1 + [1 0] x2 + [3]
                     [0 0]      [0 0]      [0]
         w() = [3]
               [0]
         cond(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                        [0 1]      [0 1]      [1]
         =(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                     [0 1]      [0 0]      [0]
         sum(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                           [0 1]      [0 0]      [0 1]      [0]
         carry(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                             [0 0]      [0 0]      [0 0]      [0]
         true() = [2]
                  [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {admit(x, .(u, .(v, .(w(), z)))) ->
           cond(=(sum(x, u, v), w()),
                .(u, .(v, .(w(), admit(carry(x, u, v), z)))))}
        Weak Trs:
          {  cond(true(), y) -> y
           , admit(x, nil()) -> nil()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component:
          {admit(x, .(u, .(v, .(w(), z)))) ->
           cond(=(sum(x, u, v), w()),
                .(u, .(v, .(w(), admit(carry(x, u, v), z)))))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(admit) = {}, Uargs(.) = {2}, Uargs(cond) = {2},
            Uargs(=) = {}, Uargs(sum) = {}, Uargs(carry) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           admit(x1, x2) = [0 0] x1 + [0 1] x2 + [3]
                           [0 0]      [0 1]      [0]
           nil() = [0]
                   [0]
           .(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                       [0 1]      [0 1]      [0]
           w() = [0]
                 [1]
           cond(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                          [0 0]      [0 1]      [0]
           =(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                       [0 0]      [0 0]      [0]
           sum(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                             [0 0]      [0 0]      [0 0]      [0]
           carry(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                               [0 0]      [0 0]      [0 0]      [0]
           true() = [0]
                    [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Weak Trs:
            {  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
             , admit(x, nil()) -> nil()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          We consider the following Problem:
          
            Weak Trs:
              {  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
               , admit(x, nil()) -> nil()}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            Empty rules are trivially bounded

Hurray, we answered YES(?,O(n^1))