Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The statement that '$x$, $y$ and $z$ are equivalent' just means

  • all of $x$, $y$ and $z$ are false, or
  • all of $x$, $y$ and $z$ are true.

Now suppose its not the case that $x$, $y$ and $z$ are equivalent. That is, suppose:

  • at least one of $x$, $y$ and $z$ is false, and
  • at least one of $x$, $y$ and $z$ is true.

Under these circumstance, we say that $x$, $y$ and $z$ are what? Inequivalent? This seems wrong, because some of them will still be equivalent.

share|improve this question
    
Related. –  Git Gud May 25 '13 at 15:05
1  
Why would you need a name for that? If I wanted to say that, I'd say just what you did: at least one of those is true, and at least one is false. I can't imagine this kind of relationship to be common enough (or rather, the need to mention it) to warrant concocting up a name for it. –  tomasz May 25 '13 at 15:31
    
@tomasz, I don't know if it shows up very often, but its interesting that its negation is its dual. Which is why I find it interesting. –  goblin May 25 '13 at 15:42
2  
How would you say if $x,y,z$ are real numbers which are not all equal? –  Asaf Karagila May 25 '13 at 16:26
    
@AsafKaragila, good question! I don't know. –  goblin May 25 '13 at 16:27

1 Answer 1

There is no "special term": We can simply say "not all of $x, y, z$ are equivalent" or "$x, y, z$ are not all equivalent."

As you note, that's not the same thing as all of them being inequivalent, because two of them might still be equivalent, and in the case of logical equivalence, we would need one pair and not all pairs logically equivalent.

Negation of the statement of mutual logical equivalence involves DeMorgan's:

We can say "$x, y, z$" are all equivalent by stating $$(x \iff y) \land (y \iff z)\tag{A}$$ and by transitivity we have also conveyed in $(A)$ that $(x \iff z)$.

And so to negate an entire conjunction, we can arrive at the disjunction of negated dijunctions $$\lnot(x \iff y) \lor \lnot(y \iff z): \tag{$\lnot $ A}$$

$\qquad\qquad\qquad \qquad\qquad\qquad$Wolfram alpha

In disjunctive normal form $(\lnot A)$ equivalent to: $$(\lnot x \land y) \lor (\lnot y \land z) \lor (x \land \lnot z)\tag{DNF}$$

And in conjunctive normal form, as you describe in your bullets: we have that $(\lnot A)$ is equivalent to $$(x \lor y\lor z) \land (\lnot x \lor \lnot y \lor \lnot z)\tag{CNF}$$

share|improve this answer
    
beautifully formatted table and nice answer! +1 –  Amzoti May 26 '13 at 1:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.