prove logic equivalence: $(a∧¬b)∨(¬c∧¬a) ≡ (a→b)→¬(c∨a)$ Hello we are asked to prove the logic equivalence of $(a∧¬b)∨(¬c∧¬a) ≡ (a→b)→¬(c∨a)$ by using the Laws from the Table of Logical Equivalences. We also have to write the name of the law used at each step (one one law at each step!)
I have tried to simplify the second part $(a→b)→¬(c∨a)$ to see if it's the same with the first part. Here is my solution so far: in the picture here.
But finally I got $(a∨¬b)∨(¬c∨¬a)$, it seems very different than the first part $(a∧¬b)∨(¬c∧¬a)$. What do I missed in my solution here?
Thank you!
 A: Let's do it step by step.

*

*Find a connection between $(a \land \lnot b)$ and $(a \rightarrow b)$. If we make a truth table, we get
$$\begin{array}{c|c|c|}
& a & b & {\lnot b} & a \land \lnot b & a \rightarrow b \\ \hline & F & F & T & F & T
\\ & F & T & F & F & T
\\ & T & F & T & T & F
\\ & T & T & F & F & T
\end{array}$$
You can see that $(a \land \lnot b)$ = $\lnot(a \rightarrow b)$ as mentioned by @Mauro ALLEGRANZA


*Using the $1.$ De Morgan's law, we have $\lnot c \land \lnot a = \lnot(c \lor a)$.

If we insert this in into your logic equivalence, we get $$ \lnot(a \rightarrow b) \lor \lnot(c \lor a) \equiv (a \rightarrow b) \rightarrow \lnot(c \lor a) $$


*Make a "variable" change. Let $\lnot(a \rightarrow b) = \lnot d$ and $\lnot(c \lor a) = \lnot e$. Then, your logic equivalence will be $$\lnot d \lor \lnot e \equiv d \rightarrow \lnot e$$
Let's make another truth table which looks as follows:
$$\begin{array}{c|c|c|}
& d & e & \lnot d & \lnot e & \lnot d \lor \lnot e & d \rightarrow \lnot e \\ \hline & F & F & T & T & T & T 
\\ & F & T & T & F & T & T
\\ & T & F & F & T & T & T
\\ & T & T & F & F & F & F
\end{array}$$
The truth table shows that indeed $$\lnot d \lor \lnot e \equiv d \rightarrow \lnot e$$ holds.


*Switch $d$ and $e$ with the original variables, so $\lnot d = \lnot (a \rightarrow b) = (a \land \lnot b)$ and $\lnot e = \lnot (c \lor a) = (\lnot c \land \lnot a)$ and you have $$(a \land \lnot b) \lor (\lnot c \land \lnot a) \equiv (a \rightarrow b) \rightarrow \lnot (c \lor a)$$ proven.

A: Are truth tables required?
$\lnot(p\implies q) \equiv p \land \lnot q$ is Negation of the Conditional.
The negation of that is $(p\implies q) \equiv q \lor \lnot p$
$(a \land \lnot b) \lor (\lnot c \land \lnot a)\equiv (a \land \lnot b )\lor \lnot (c \lor a)$ by DeMorgan's Law on the right most term.
$(a \land \lnot b) \lor (\lnot c \land \lnot a)\equiv \lnot(a \implies b) \lor \lnot (c \lor a)$ by Negation of the Conditional for the a,b term.
$(a \land \lnot b) \lor (\lnot c \land \lnot a)\equiv (a \implies b) \implies \lnot (c \lor a)$ *
*Let $q=\lnot(c \lor a)$ and $p=(a \implies b )$ Then we have a statement in the form $\lnot p \lor q$ allowing us to use the negation of Negation of the Conditional, or equivalently, the re-expression of a conditional in terms of negation and or statement.
