(A&B)->~A

Use the principle p->q is equivalent to ~pVq

~(A&B)V~A

Use deMorgan's law:  ~(p&q) is equivalent to ~pV~q

(~AV~B)V~A

Use the commutative law:  pVq is equivalent to qVp

(~BV~A)V~A

Use the associative law:  (pVq)Vr is equivalent to pV(qVr)

~BV(~AV~A)

Use the idempotent law:   pVp is equivalent to p

~BV~A

Looks like the argument is not valid.

Edwin