Logical Equivalence Derivation (Propositonal Logic)

Solution 1:

Your problem can be solved very easily using the Consensus Theorem, which states that:

$(X \lor Y) \land (\neg Y \lor Z) \land (X \lor Z) \Leftrightarrow (X \lor Y) \land (\neg Y \lor Z)$

(this law is most easily remembered by going from right to left: with the $(X \lor Y)$ term containing a $Y$, and the $(\neg Y \lor Z)$ term containing a $\neg Y$, you can put together the other parts ($X$ and $Z$) into a new term $(X \lor Z)$ and just add that as a third term.

So with that:

$$(A\land\lnot B)\lor(B\land\lnot C)$$

$$\overset{Distribution}{\Leftrightarrow}$$

$$(A\lor B)\land(A\lor\lnot C)\land(\lnot B\lor B)\land(\lnot B\lor\lnot C)$$

$$\overset{Tautology}{\Leftrightarrow}$$

$$(A\lor B)\land(A\lor\lnot C)\land(\lnot B\lor\lnot C)$$

$$\overset{Consensus}{\Leftrightarrow}$$

$$(A\lor B)\land(\lnot B\lor\lnot C)$$

$$\overset{DeMorgan}{\Leftrightarrow}$$

$$(A\lor B)\land \lnot (B\land C)$$

If you are not allowed to use Consensus in one step, please know that it is easily derived as follows:

$$(X \lor Y) \land (\neg Y \lor Z) \land (X \lor Z)$$

$$\overset{Adjacency}{\Leftrightarrow}$$

$$(X \lor Y) \land (\neg Y \lor Z) \land (X \lor Z \lor Y) \land (X \lor Z \lor \neg Y)$$

$$\overset{Absorption \ x \ 2}{\Leftrightarrow}$$

$$(X \lor Y) \land (\neg Y \lor Z)$$

Solution 2:

\begin{align} (A\land\lnot B)\lor(B\land\lnot C)&=(A\lor B)\land(A\lor\lnot C)\land(\lnot B\lor B)\land(\lnot B\lor\lnot C)\\ &=(A\lor B)\land(A\lor\lnot C)\land(\lnot B\lor\lnot C)\\ &=((A\lor B)\land(A\lor\lnot C))\land((A\lor\lnot C)\land(\lnot B\lor\lnot C))\\ &=(A\lor(B\land\lnot C))\land((A\land\lnot B)\lor\lnot C)\\ &=(A\land(A\land\lnot B))\lor(A\land\lnot C)\\ &\quad\lor((B\land\lnot C)\land(A\land\lnot B))\lor((B\land\lnot C)\land\lnot C)\\ &=(A\land\lnot B)\lor(A\land\lnot C)\lor(B\land\lnot C)\\ &=(A\land\lnot B)\lor(B\land\lnot B)\lor(A\land\lnot C)\lor(B\land\lnot C)\\ &=((A\lor B)\land\lnot B)\lor((A\lor B)\land\lnot C)\\ &=(A\lor B)\land(\lnot B\lor\lnot C)\\ &=(A\lor B)\land\lnot(B\land C) \end{align}