Logic - Will a second parameter value inherit negation if the first parameter is false? Will a second parameter value inherit negation if the first parameter is false?
Like:
(~A & B) → X

Is B false? Would it translate to:
if A and B is false, then X

or
if A is false and B, then X

I'm very new to propositional calculus, and had a hard time phrasing the question, but any help would be much appreciated!
 A: To answer your question, $(\sim A \& B) \to X$ always means "If $A$ is false and $B$ is true, then $X$ is true".  It never means "If $A$ is false and $B$ is false then $X$ is true".
The jargon here is this: You are asking about the "relative precedence" of $\sim$ and $\&$.  Whenever you have two operators, say $\sim$ and $\&$, you can write expressions involving them with complete parentheses:


*

*$((\sim A) \& B)$

*$\sim(A\& B)$


These mean different things, and both are completely unambiguous.  But we can also omit some of the parentheses and leave it up to convention which of the fully-parenthesized versions is meant.  In this case you are asking whether $$(\sim A\& B)$$
means (1) or (2). If the effect of $\sim$ attaches only to $A$, as in (1), we say that $\sim$ has "higher precedence" than $\&$; if instead the effect of $\&$ attaches directly to $A$ and $B$, leaving the $\sim$ to apply to the larger $A\&B$ expression, as in (2), we say that $\&$ has higher precedence than $\sim$.
The universal convention is that $\sim$ has higher precedence than $\&$, so your expression means (1), not (2).
This is analogous to the way that $1 \times 2 + 3$ always means $(1 \times 2)+ 3$, not $1 \times (2 + 3)$.
A: No, because $${\sim}A \wedge B \not\equiv {\sim}A \wedge {\sim}B.$$ The latter translation is correct i.e. ${\sim}A \wedge B \Rightarrow X$ means "If $A$ is false and $B$ is true, then $X$ is true." Do not confuse this with the De Morgan's laws:
$${\sim}(A \wedge B) \equiv\sim A \vee {\sim}B$$
$${\sim}(A \vee B) \equiv \sim A \wedge {\sim}B.$$
