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Answer to Question equation-2

For reference, here is the equation that we must show is false. 
  isPrefix(h:t, L) = isPrefix(t, L)
There are many counterexamples. How about h = 1, t = [2], L = [2,2]. The left-hand side of the equation is 
  isPrefix(1:[2], [2,2]) = isPrefix([1,2], [2,2])
                         = false                     since [1,2] is not a prefix of [2,2].
The right-hand side of the equation is 
  isPrefix([2], [2,2]) = true                        since [2] is a prefix of [2,2].
The two sides are not the same for this example.