YES

Problem:
 del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
 f(true(),x,y,z) -> del(.(y,z))
 f(false(),x,y,z) -> .(x,del(.(y,z)))
 =(nil(),nil()) -> true()
 =(.(x,y),nil()) -> false()
 =(nil(),.(y,z)) -> false()
 =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
                   [1 0]     [1 0]     [0]
   [and](x0, x1) = [1 0]x0 + [0 0]x1 + [1],
   
         [0]
   [v] = [0],
   
         [3]
   [u] = [0],
   
           [0]
   [nil] = [0],
   
             [2]
   [false] = [2],
   
            [1]
   [true] = [1],
   
                              [1 0]     [2 0]     [3 0]     [3]
   [f](x0, x1, x2, x3) = x0 + [0 0]x1 + [3 0]x2 + [3 0]x3 + [2],
   
                 [1 0]     [1 0]     [1]
   [=](x0, x1) = [2 0]x0 + [0 0]x1 + [3],
   
               [2 1]     [2]
   [del](x0) = [3 0]x0 + [0],
   
                 [1 0]     [1 0]     [1]
   [.](x0, x1) = [0 0]x0 + [1 0]x1 + [0]
  orientation:
                      [2 0]    [3 0]    [3 0]    [7]    [2 0]    [3 0]    [3 0]    [4]                  
   del(.(x,.(y,z))) = [3 0]x + [3 0]y + [3 0]z + [6] >= [2 0]x + [3 0]y + [3 0]z + [5] = f(=(x,y),x,y,z)
   
                     [1 0]    [2 0]    [3 0]    [4]    [2 0]    [3 0]    [4]              
   f(true(),x,y,z) = [0 0]x + [3 0]y + [3 0]z + [3] >= [3 0]y + [3 0]z + [3] = del(.(y,z))
   
                      [1 0]    [2 0]    [3 0]    [5]    [1 0]    [2 0]    [3 0]    [5]                   
   f(false(),x,y,z) = [0 0]x + [3 0]y + [3 0]z + [4] >= [0 0]x + [2 0]y + [3 0]z + [4] = .(x,del(.(y,z)))
   
                    [1]    [1]         
   =(nil(),nil()) = [3] >= [1] = true()
   
                     [1 0]    [1 0]    [2]    [2]          
   =(.(x,y),nil()) = [2 0]x + [2 0]y + [5] >= [2] = false()
   
                     [1 0]    [1 0]    [2]    [2]          
   =(nil(),.(y,z)) = [0 0]y + [0 0]z + [3] >= [2] = false()
   
                          [1 0]    [1 0]    [6]    [1 0]    [1 0]    [5]                         
   =(.(x,y),.(u(),v())) = [2 0]x + [2 0]y + [5] >= [1 0]x + [0 0]y + [5] = and(=(x,u()),=(y,v()))
  problem:
   f(true(),x,y,z) -> del(.(y,z))
   f(false(),x,y,z) -> .(x,del(.(y,z)))
   =(nil(),nil()) -> true()
   =(.(x,y),nil()) -> false()
   =(nil(),.(y,z)) -> false()
  Matrix Interpretation Processor: dim=3
   
   interpretation:
            [0]
    [nil] = [1]
            [0],
    
              [0]
    [false] = [0]
              [0],
    
             [0]
    [true] = [0]
             [0],
    
                          [1 0 0]     [1 0 0]     [1 0 0]     [1 0 0]     [1]
    [f](x0, x1, x2, x3) = [0 0 0]x0 + [0 0 0]x1 + [0 0 0]x2 + [0 0 0]x3 + [0]
                          [0 0 0]     [0 0 0]     [0 0 0]     [0 0 0]     [1],
    
                  [1 1 1]     [1 0 0]  
    [=](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                  [0 0 0]     [0 0 0]  ,
    
                [1 0 0]  
    [del](x0) = [0 0 0]x0
                [0 0 1]  ,
    
                  [1 0 0]     [1 0 0]     [0]
    [.](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                  [0 0 0]     [0 0 0]     [1]
   orientation:
                      [1 0 0]    [1 0 0]    [1 0 0]    [1]    [1 0 0]    [1 0 0]    [0]              
    f(true(),x,y,z) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] >= [0 0 0]y + [0 0 0]z + [0] = del(.(y,z))
                      [0 0 0]    [0 0 0]    [0 0 0]    [1]    [0 0 0]    [0 0 0]    [1]              
    
                       [1 0 0]    [1 0 0]    [1 0 0]    [1]    [1 0 0]    [1 0 0]    [1 0 0]    [0]                   
    f(false(),x,y,z) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] >= [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] = .(x,del(.(y,z)))
                       [0 0 0]    [0 0 0]    [0 0 0]    [1]    [0 0 0]    [0 0 0]    [0 0 0]    [1]                   
    
                     [1]    [0]         
    =(nil(),nil()) = [0] >= [0] = true()
                     [0]    [0]         
    
                      [1 0 0]    [1 0 0]    [1]    [0]          
    =(.(x,y),nil()) = [0 0 0]x + [0 0 0]y + [0] >= [0] = false()
                      [0 0 0]    [0 0 0]    [0]    [0]          
    
                      [1 0 0]    [1 0 0]    [1]    [0]          
    =(nil(),.(y,z)) = [0 0 0]y + [0 0 0]z + [0] >= [0] = false()
                      [0 0 0]    [0 0 0]    [0]    [0]          
   problem:
    
   Qed