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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


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!

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I understand you had a hard time phrasing the question, but it makes no sense to me. I suggest that instead of trying to phrase it with jargon, you try to explain more naturally what you are trying to find out, perhaps with examples. – MJD Dec 10 '12 at 0:00
I added an example. – user52428 Dec 10 '12 at 0:03
I am also studying programming logic for computer science, however I guess the jargon doesn't really cross over as I thought it would. The "parameter values" being A and B, a parameter being a string enclosed in brackets, then negation being ~ or !. – user52428 Dec 10 '12 at 0:04
In terms of translating, the second option is the correct translation: if (A is false and B), then X. – amWhy Dec 10 '12 at 0:05
Oh, okay, thanks! – user52428 Dec 10 '12 at 0:07

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:

  1. $((\sim A) \& B)$
  2. $\sim(A\& B)$

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 $1 \times (2 + 3)$.

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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: $${\sim}(A \wedge B) \equiv\sim A \vee {\sim}B$$ $${\sim}(A \vee B) \equiv \sim A \wedge {\sim}B.$$

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