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I have always thought of implication in the way it is shown here in wikipedia: Contraposition article

When I think of the first part of the implication as the violet (sub-)set A and the second part of the implication as the blue (super-)set B then implication makes sense to me.

  • Something that isn't contained within A might not be contained in B too. => Right.

  • Something that isn't contained within A can still be contained in B. => Right.

  • Something is contained in A but not in B doesn't make sense. So therefore implication becomes false.

  • Something that is contained in A is contained in B as well. => Right.

That way of thinking has always made sense to me.

Now I have to learn proof by contradiction.

I have got this table as part of my study documents.

Truth table

I try to understand but no chance.

I mean especially the third row.

If I invers p and q. So that p (which is the subset in my thinking) becomes true and q (which is the superset) becomes false: How can this be true?

Wouldn't it say then:

"It isn't in the superset (q == false) but it is in the subset (p)"?

Because implication from false to true becomes true implication the result is true.

But that doesn't make sense to me.

How can something be part of a subset but not part of the superset?

Can anybody help me to understand how this table works?

Respectively: What do I wrong concerning my thinking? Where's my error?

mizech
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  • You may find this helpful: http://math.stackexchange.com/questions/1551320/understanding-vacuously-true-truth-table/1551525#1551525 – Dan Christensen Apr 14 '17 at 14:05
  • Indeed very interesting. Thanks a lot. :) – mizech Apr 14 '17 at 16:06
  • @DanChristensen I think the link may confuse things. If there were (but there isn't) a column in the truth-table labeled $(p \leftrightarrow q)$, then we would have the last comment in the answer/question you linked to. As I said, that's not a column here. Instead, we have the tautology, expressing the truth values of the biconditional involving equivalent statements $(p\to q) \leftrightarrow(\lnot q \to \lnot p)$. And as we can see, it expresses a tautology. Whenever $p\to q$ is true, so is $\lnot q \to \lnot p$ true. Whenever, $p\to q$ is false, so is $\lnot q \to \lnot p$. – amWhy Apr 14 '17 at 16:25
  • Looking at your answer (which you linked to here), is isn't even relevant here. – amWhy Apr 14 '17 at 16:27
  • It seemed to me that the OP was confused about the notion of material implication among other things. I thought my link might help him to solve his problem at least in part. It is comment after all, not an answer. – Dan Christensen Apr 14 '17 at 17:40

1 Answers1

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Your translation to "It isn't in the superset but it is in the subset" is erroneous. To complement the logic, you must complement the sets - BOTH of them. Switch 'inside' with 'outside'. I.e:

p --> q is "Whatever is INside p, is INside q"

translates to

-q --> -p is "Whatever is OUTside q, is OUTside p"

PMar
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