Top/Data structures/Lists/Equations/Answer-equation-5

Answer to Question equation-5

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([2,3,4], [2,4,3]) = isPrefix(2:[3,4], 2:[4,3])
                             = 2 == 2  and  isPrefix([3,4], [4,3])    by (isPrefix.3)
                             = true  and  isPrefix([3,4], [4,3])
                             = isPrefix([3,4], [4,3])
                             = isPrefix(3:[4], 4:[3])
                             = 3 ==4  and  isPrefix([4], [3])         by (isPrefix.3)
                             = false  and  isPrefix([4], [3])
                             = false