Top/Data structures/Lists/Equations/Answer-equation-6

Answer to Question equation-6

Here are the equations for reference. 
  (isPrefix.1)   isPrefix([], B)        = true
  (isPrefix.2)   isPrefix(A, [])        = false    (when A is not [])
  (isPrefix.3)   isPrefix(h1:t1, h2:t2) = h1 == h2  and  isPrefix(t1, t2)
Evaluation is as follows. 
  isPrefix([2,3], [2]) = isPrefix(2:[3], 2:[])
                       = 2 == 2  and  isPrefix([3], [])    by (isPrefix.3)
                       = true  and  isPrefix([3], [])
                       = isPrefix([3], [])
                       = isPrefix(3:[], [])
                       = false                             by (isPrefix.2)