Not understanding how to do this logic question Let $A, B$ be propositional formulas. Demonstrate that if there exists a propositional formula $C$ such that $A$ is a logical consequence of $C$ and $B$ is a logical consequence of $\neg C$, then formula $A \lor B$ is a tautology. 
 A: First think about it informally. If $C$ is true, then $A$ is true, and if $C$ is false, then $B$ is true; since $C$ must be true or false, $A$ or $B$ must be true. What we’ve used here besides the given hypotheses is the fact that $C\lor\neg C$ is a tautology. The problem now is to convert this into a more formal argument. The details will depend on just what formalism you’re using. I’ll use $\equiv$ to indicate that two propositional expressions are logically equivalent, and $\Rightarrow$ to indicate that one entails another. I’ll start with the two givens, which are essentially that $C\to A$ and $\neg C\to B$ are true, and the tautology $C\lor\neg C$, and use fairly standard manipulations to arrive $A\lor B$ in a way that more or less mimics the informal argument.
$$\begin{align*}
(C\lor\neg C)\land(C\to A)\land(\neg C\to B)&\equiv\Big(\big(C\land(C\to A)\big)\lor\big(\neg C\land(C\to A)\big)\Big)\\
&\qquad\land(\neg C\to B)\\
&\Rightarrow\Big(A\lor\big(\neg C\land(C\to A)\big)\Big)\land(\neg C\to B)\\
&\equiv\big(A\land(\neg C\to B)\big)\lor\Big(\big(\neg C\land(C\to A)\big)\land(\neg C\to B)\Big)\\
&\Rightarrow A\lor\Big(\big(\neg C\land(\neg C\to B)\big)\land(C\to A)\Big)\\
&\Rightarrow A\lor\big(B\land(C\to A)\big)\\
&\equiv(A\lor B)\land\big(A\lor(C\to A)\big)\\
&\Rightarrow A\lor B
\end{align*}$$
A: You can do it with a truth-table:
   C   |   A   |   B   | A v B 
-------|-------|-------|-------
   0   |   x   |   1   |   1   
-------|-------|-------|-------
   1   |   1   |   x   |   1   

As you can see, the expression $A \vee B$ is always true.
A: The following are tautologies:
$$
\begin{align}
(C \to A) &\iff (\neg A \to \neg C) \tag{1} \\
(\neg A \to B) &\iff (A \vee B) \tag{2}
\end{align}
$$
so by (1) the premises are equivalent to:
$$
\begin{align}
&(\neg A \to \neg C) \tag{i} \\
&(\neg C \to B) \tag{ii} \\
&(A \vee B) \tag{conclusion} \\
\end{align}
$$
From (i) an (ii) we can conclude:
$$
(\neg A \to B)$$
which by (2) is equivalent to:
$$
(A \vee B) \text{.}
$$
