and the negation of the loop condition,
{ x < y}, their conjunction is
{input1 = div * y + x and x < y}
which does NOT imply,
{input1 = div * input2 + x and
0 =< x < input2}
We need to add {y = input2} and
{ x >= o}.
So we formulate a stronger invariant:
L2: {input1 = div * y + x and
x >= 0 and y = input2}