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When proving that $ab = 0 \implies a = 0 \,\mbox{ or }\,b = 0$ for members $a$ and $b$ of a field, I used an argument like

  1. Suppose $ab = 0$ and $a \ne 0$ ... then $b = 0$.
  2. Now suppose $ab = 0$ and $b \ne 0$ ... then $a = 0$.
  3. Therefore, if $ab = 0$, then $a = 0$ or $b = 0$.

The general form of that argument would, as far as I can tell, be

$$ (p \land \lnot q \to r) \land (p \land \lnot r \to q) \to (p \to q \lor r) $$

Is that general form indeed a valid argument? How can I know for sure? (Is there a "for sure"?)

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    If the elements $a,b$ belong to a field, then the condition is already met, as one of the definitive properties of a field is that a field satisfies all the properties of an integral domain, including that of having no non-zero zero divisors... But I'm sure this isn't what you're looking for ;)2012-09-04
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    Just to be clear, building on Ed's comment, are you studying algebra or are you studying logic? If you're studying logic, this is fine. If you're studying algebra, but just trying to use logic to solve a problem, you're making things too complicated, as I explain in my answer below.2012-09-04
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    @Graphth: I'm studying logic (I think). I just included the algebra to provide motivation for that expression I came up with.2012-09-04
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    @Blue: Then it looks like what you're looking for is _proof systems_ in _formal logic_, which is concerned with syntactic manipulation to prove things. There are several different styles of doing this. In simple cases like this most of them are scarcely more enlightening than truth tables, but they become more interesting when you add rules for quantifiers. For a start take a look at [Hilbert system](http://en.wikipedia.org/wiki/Hilbert_system), [natural deduction](http://en.wikipedia.org/wiki/Natural_deduction), and [Sequent calculus](http://en.wikipedia.org/wiki/Sequent_calculus) on WP.2012-09-04

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