Can I deduce $\lnot W, \lnot U$ from $\lnot W \land \lnot U$? I have to prove this by contradiction for school :
$(X \lor Y) \implies M,\space M\implies (W \lor Z),\space Z \implies U, ¬(W\lor U)$ together, imply $¬Y $  
I start by doing this :  
$(X\lor Y) \implies M   \space $ Premise
$M\implies (W \lor Z)$  Premise  
$Z\implies U \space \space\space\space\space\space\space\space\space\space\space\space\space $  Premise  
$¬(W \lor U) \space\space\space\space\space\space\space\space\space\space\space\space\space$  Premise  
$¬(¬Y)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$  Contradiction of the conclusion  
$ Y \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$  Double negation  
$¬W \land ¬U\space\space\space\space\space\space\space\space\space\space\space\space\space$  De Morgan  
I would like to know if I can do :
$¬W\land ¬U$
$¬W$
$¬U$  
I would like to know if it's true so I can continue.
Thanks in advance for your help.
 A: The proof is:


*

*(X∨Y)⟹M    (premise)

*M⟹(W∨Z)    (premise)

*Z⟹U        (premise)

*¬(W∨U)       (premise)

*¬W∧¬U       (de Morgan, 4)

*¬U          (simplification, 5)

*¬W          (simplification, 5)

*¬Z          (Modus Tollens, 3, 6)

*¬W∧¬Z       (conjuntion, 7, 8)

*¬(W∨Z)      (de Morgan, 9)

*¬M         (Modus Tollens, 2, 10)

*¬(X∨Y)     (Modus Tollens, 1, 11)

*¬X∧¬Y      (de Morgan, 12)

*¬Y         (simplification, 13)


And, if you wish, by Contradiction:


*

*(X∨Y)⟹M    (premise)

*M⟹(W∨Z)    (premise)

*Z⟹U        (premise)

*¬(W∨U)       (premise)

*Y            (sup. for contradiction)

*¬W∧¬U       (de Morgan, 4)

*¬U          (simplification, 6)

*¬W          (simplification, 6)

*¬Z          (Modus Tollens, 3, 7)

*X∨Y          (adition, 5)

*M            (modus ponnens, 1, 10)

*W∨Z          (modus ponnens, 2, 11)

*Z            (disjunctive syllogism, 12, 8)

*¬Y           (by suppose 5, contradiction in 9, 13)

A: Assuming you want to prove $P \implies Q$, then:
A proof by contradiction would be $(P \land \lnot Q) \implies (R \land \lnot R)$.  We can't have both $R$ and $\lnot R$ true at the same time, hence the contradiction.
Related:  A proof by contrapositive would be $\lnot Q \implies \lnot P$.
