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.