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I've got a question about the use of implications during the process of proving something. I will get more specific. I tried to solve the following problem:

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The solution is straight forward. Show that L satisfies all axioms of a poset under the given conditions. I've got stuck proving that $a \le b \wedge b\le c \implies a \le c$.

Here's my attempt to present a solution.

$a \le b \wedge b\le c$

$\iff (\ a = b \vee a < b)\ \wedge (\ b = c \vee b < c)\ $

$\iff [\ (\ a = b \vee a < b)\ \wedge (\ b = c )\ ]\ \vee [\ (\ a = b \vee a < b)\ \wedge (\ b < c )\ ]\ $

$\iff \langle [\ (\ a = b )\ \wedge (\ b=c)\ ]\ \vee [\ (\ a < b )\ \wedge (\ b = c )\ ]\ \rangle \vee \langle [\ (\ a = b)\ \wedge (\ b < c )\ ]\ \vee [\ (\ a < c)\ \wedge (b < c) ]\ \rangle $

$\iff [\ (\ a = b )\ \wedge (\ b=c)\ ]\ \vee [\ (\ a < c)\ \wedge (b < c) ] \vee [\ (\ a < b )\ \wedge (\ b = c )\ ]\ \vee [\ (\ a = b)\ \wedge (\ b < c )\ ]\ \ $

$\implies [\ (\ a = b )\ \wedge (\ b=c)\ ]\ \vee [\ (\ a < c)\ \wedge (b < c) ] \implies (a = c) \wedge (a

Now, while this seems to lead to a correct solution, I am very unsure about the first implication I've made by omitting two of the []'s. Can somebody tell meif this is correct? And if not, what would be the correct formal proof?

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    Feel free to edit my question title. I wasn't sure how I should phrase it. – 2017-02-21
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    Are you not allowed to use words in your proof? – 2017-02-21
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    @David This textbook didn't say explicitly say that I couldn't use them. But it doesn't matter really. The real question is, if it is correct or not to make such an implication as I did. – 2017-02-21
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    I don't agree that it doesn't matter. One of the essential things about a proof is to communicate an argument to someone else. A clearly written proof with words is, nearly always, way more comprehensible than a wordless string of symbols. If you frame your proof this way: "Let $a\le b$ and $b\le c$. Case I: $a=b$... Case II: $b=c$... Case III: neither of the above..." then I think you will find it very easy to complete the proof by filling in the dots, **and moreover**, answering your original question, very easy to be sure that your logic is correct. – 2017-02-21

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Your next-to-last line reads: $$[(a=b) \wedge (b=c) ] \vee [(a

What is left is to show that all four disjuncts imply $a \leq c$.

  • $[(a=b) \wedge (b=c)]$ implies $a=c$ by transitivity of equality; $a=c$, in turn, implies $a \leq c$.
  • $[(a
  • $[(a
  • $[(a=b) \wedge (b

As to the question whether you can omit two disjuncts, the answer, if I understand the question, is no. You cannot focus on the cases that support the conclusion you are trying to prove and ignore the others.

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    Believe it or not, I was actually able to prove reflexivity and antisymmetry on my own. I got stuck at the transitivity property. – 2017-02-21
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    Believe it or not, I was fully convinced I had read the statement of antisymmetry. – 2017-02-21
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    Thank you, I wasn't aware that $[\ (\ a \implies d )\ \wedge (\ b \implies d )\ ]\ \iff [\ (\ a \vee b )\ \implies d ]\ $ holds. – 2017-02-22