What is equivalent that logical function?

Question

(A+B) => (B XOR C)

=> is Implication

Options:

BC+!A+!B+!C

!B!C+ABC

B!C+A+!BC

!BC+!ABC

BC+A!B!C

! is NOT

Please tell me how it is solved. I do not think that this requires knowledge of discrete mathematics. But I did not find any simple options

Answer 1

$$(A+B) \Rightarrow (B \ XOR \ C) = \text{ (Rewrite XOR)}$$

$$(A+B) \Rightarrow (B!C + !BC) = \text{ (Rewrite} \Rightarrow )$$

$$!(A + B)+B!C + !BC = \text{ (DeMorgan)}$$

$$!A!B+B!C+!BC$$

.... which is not any of the options you are providing ...

$BC + !A +!B+!C=1$ for $A = 0$ and $B=C=1$, but then $!A!B+B!C+!BC = 0$

$!B!C+ABC=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$

$B!C + A +!BC=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$

$!BC+!ABC=1$ for $A = 0$ and $B=C=1$, but then $!A!B+B!C+!BC = 0$

$BC+A!B!C=1$ for $A=B=C=1$, but then $!A!B+B!C+!BC = 0$

Are you sure you wrote this all correctly from your exercise?

EDIT

Aha! So the original is:

$$!((A+B) \Rightarrow (B \ XOR \ C))$$

So then you get:

$$!((A+B) \Rightarrow (B \ XOR \ C)) = \text{(Rewrite XOR)}$$

$$!((A+B) \Rightarrow (B!C + !BC)) = \text{ (Rewrite} \Rightarrow )$$ $$!(!(A + B) + B!C + !BC) = \text{ (DeMorgan)}$$

$$(A + B)!(B!C + !BC) = \text{ (DeMorgan)}$$

$$(A + B)(!(B!C)!(!BC)) = \text{ (DeMorgan)}$$

$$(A + B)(!B+C)(B+!C) = \text{ (Distribution)}$$

$$A!BB+A!B!C+ACB+AC!C+B!BB+B!B!C+BCB+BC!C = \text{ (Complement)}$$

$$A0+A!B!C+ACB+A0+0B+0!C+BCB+B0 = \text{ (Annihilation)}$$

$$0+A!B!C+ACB+0+0+0+BCB+0 = \text{ (Identity)}$$

$$A!B!C+ACB+BCB = \text{ (Commutation)}$$

$$A!B!C+ABC+BBC = \text{ (Idempotence)}$$

$$A!B!C+ABC+BC = \text{ (Absorption)}$$

$$A!B!C+BC = \text{ (Commutation)}$$

$$BC+A!B!C$$