YES

Problem:
 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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                        
   [a](x0, x1) = x0 + x1
                        ,
   
         [0]
   [t] = [0]
         [0],
   
                       [1]
   [lambda](x0) = x0 + [0]
                       [0],
   
         [0]
   [1] = [0]
         [0]
  orientation:
                            [1]        [1]                   
   a(lambda(x),y) = x + y + [0] >= x + [0] = lambda(a(x,1()))
                            [0]        [0]                   
   
                            [1]            [1]                        
   a(lambda(x),y) = x + y + [0] >= x + y + [0] = lambda(a(x,a(y,t())))
                            [0]            [0]                        
   
                                                     
   a(a(x,y),z) = x + y + z >= x + y + z = a(x,a(y,z))
                                                     
   
                   [1]         
   lambda(x) = x + [0] >= x = x
                   [0]         
   
                          
   a(x,y) = x + y >= x = x
                          
   
                          
   a(x,y) = x + y >= y = y
                          
  problem:
   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))
   a(x,y) -> x
   a(x,y) -> y
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(lambda) = 1
    w(t) = w(1) = w(a) = 0
   status function:
    st(t) = st(1) = []
    st(a) = [0, 1]
    st(lambda) = [0]
   precedence:
    a > t ~ 1 ~ lambda
   problem:
    
   Qed