We consider the following Problem:

  Strict Trs:
    {  a(lambda(x), y) -> lambda(a(x, 1()))
     , a(lambda(x), y) -> lambda(a(x, a(y, t())))
     , a(a(x, y), z) -> a(x, a(y, z))
     , lambda(x) -> x
     , a(x, y) -> x
     , a(x, y) -> y}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  Arguments of following rules are not normal-forms:
  {  a(a(x, y), z) -> a(x, a(y, z))
   , a(lambda(x), y) -> lambda(a(x, 1()))
   , a(lambda(x), y) -> lambda(a(x, a(y, t())))}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  lambda(x) -> x
       , a(x, y) -> x
       , a(x, 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: {a(x, y) -> y}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(a) = {}, Uargs(lambda) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       a(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
                   [0 0]      [0 1]      [0]
       lambda(x1) = [1 0] x1 + [0]
                    [0 0]      [0]
       1() = [0]
             [0]
       t() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  lambda(x) -> x
         , a(x, y) -> x}
      Weak Trs: {a(x, 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: {a(x, y) -> x}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(a) = {}, Uargs(lambda) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         a(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
                     [0 1]      [0 1]      [0]
         lambda(x1) = [1 0] x1 + [0]
                      [0 0]      [0]
         1() = [0]
               [0]
         t() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {lambda(x) -> x}
        Weak Trs:
          {  a(x, y) -> x
           , a(x, 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: {lambda(x) -> x}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(a) = {}, Uargs(lambda) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           a(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                       [0 1]      [0 1]      [0]
           lambda(x1) = [1 0] x1 + [1]
                        [0 1]      [0]
           1() = [0]
                 [0]
           t() = [0]
                 [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Weak Trs:
            {  lambda(x) -> x
             , a(x, y) -> x
             , a(x, y) -> y}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          We consider the following Problem:
          
            Weak Trs:
              {  lambda(x) -> x
               , a(x, y) -> x
               , a(x, y) -> y}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            Empty rules are trivially bounded

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