Seems like assuming the negation would give you a contradiction:
use ~ for 'negation', and & for 'and'; then your negated assumption becomes:
From which you get:
~B&(~A\/B) (By & elimination)
I think you can show from here that your (negated) assumption leads to a contradiction, by arguing by cases, from ~A\/B, and showing neither case is possible.
Now, ~A\/B follows using a second & elimination. Show neither ~A nor B is possible
from the premises in your negated statement. Then from ~A\/B and ~(~A) and ~B,
a contradiction to your negated assumption follows.
Edit: this may be much simpler, tho yoyo's answer may be better than
mine in that he gives a direct proof, and mine is by contradiction:
Assume the negation of your statement:
ii)Conclude A, by detachment.
iii)Conclude ~A by detachment inside of parenthesis.
iv)Negation of 'i)' follows by contradiction A&~A
Where detachment--more precisely, &-detachment-- is the rule that allows us to conclude either A, or B, from a statement A&B. To show it is a valid rule, you can either use a truth table to show it is a tautology/logical truth, derive it from the empty set of premises (this is the definition of theorem I am more familiar with), or,equivalently, show that the negation of either of :(A&B-->A) or of (A&B-->A), is a contradiction.For the second approach, assume A&B, and just conclude A (seems tautological, but it works; many of these arguments in sentence logic seem tautological anyway), or transform the implication A&B->A into the equivalent statement ~(A&B)\/A == ~A\/~B\/A== ~A\/A\/B==(~A\/A)\/B (I'm using here the result that A->B is truth-functionally equiv. to ~A\/B), which is a tautology.