Conjunctive Normal Form to Disjunctive Normal Form My question is this...
Convert: ((A->B)&(~A->C)) into ((A&B)|(~A&C)) using the natural deduction system
My working so far is:
(A->B) by simplification
(~A->C) by commutation and simplification
Then, using definition of implication on both of them, i get:
(~A|B)
(A|C)
Then, by the law of conjunction, i arrive at:
(~A|B)&(A|C)
I get stuck after this. By using a few distribution laws i managed to get it looking like this: 
((A&~A)|(A&B)) | ((~A|B)&C)
where (A&~A) must be false but I also can't seem to find a rule to remove the (A&~A) part. Help please.
Also, The question also hints that a conditional proof will come in handy, but i just can't see where it should be applied because conditional proofs always end up with "assumption IMPLIES conclusion" but there is no "->" in the final answer. 
 A: With Natural Deduction :
1) $(A \to B) \land (\lnot A \to C)$ --- premise
2) $(A \to B)$ --- from 1 by $\land$-elimination (i.e. simplification)
3) $(\lnot A \to C)$ --- from 1 by $\land$-elimination (i.e. simplification)
4) $A \lor \lnot A$ --- Excluded Middle : is necessary for classical propositional calculus
5) $A$ --- assumed [a]
6) $B$ --- from 5) and 2) by $\to$-elimination
7) $A \land B$ --- from 5) and 6) by $\land$-introduction
8) $(A \land B) \lor (\lnot A \land C)$ --- from 7) by $\lor$-introduction
9) $\lnot A$ --- assumed [b]
10) $C$ from 9) and 3) by $\to$-elimination
11-12) $(A \land B) \lor (\lnot A \land C)$ --- from 9) and 10) by by $\land$-introduction followed by $\lor$-introduction, as above
13) $(A \land B) \lor (\lnot A \land C)$ --- from 4), 5) 9) and 12) by $\lor$-elimination, discharging [a] and [b]

14) $[(A \to B) \land (\lnot A \to C)] \to [(A \land B) \lor (\lnot A \land C)]$ --- from 1) and 13) by $\to$-introduction.


In the same way, we can derive the other implication :

15) $[(A \land B) \lor (\lnot A \land C)] \to [(A \to B) \land (\lnot A \to C)]$

concluding with :


$[(A \to B) \land (\lnot A \to C)] \leftrightarrow [(A \land B) \lor (\lnot A \land C)]$ --- from 14) and 15) by $\leftrightarrow$-introduction.


