0
$\begingroup$

I have a problem of the form:

The case $x≠a$ and $y=b$ is impossible.

The case $x≠a$ and $y≠b$ is impossible.

Does the case: $x=a$ for all $y$ is the only result of this problem?

  • 2
    Yes ${}{}{}{}{}$ – 2017-01-07
  • 0
    @OpenBall: I mean for all $y$. – 2017-01-07
  • 0
    I know. ${}{}{}{}$ – 2017-01-07
  • 0
    Contraposing both you have : $\lnot (y=b) \to (x=a)$ and $(y=b) \to (x=a)$. But $\lnot (y=b) \lor (y=b)$ is a logical law; thus by [Disjunction elimination](https://en.wikipedia.org/wiki/Disjunction_elimination) it follows that : $(x=a)$. – 2017-01-07

2 Answers 2

1

Yes. Take any $y$. Assume (as a proof by Contradiction) that $x\not =a$. Now, either $y =b$ or $y \not = b$. Given $x\not =a$ either case leads to a contradiction. Hence, our assumption that $x\not =a$ must be mistaken Hence, for any $y$: $x = a$.

  • 0
    @ Bram28: I mean $x=a$ not the converse – 2017-01-07
  • 0
    @E.James Sorry, I had that just the other way around! Fixed in my Answer now. – 2017-01-07
1

De Morgan's laws.

Let p = (x=a) and q = (y=b).

Note: "~" means "not".

We are given ~(~p and q) and ~(~p and ~q).

These are equivalent to (p or ~q) and (p or q).

By De Morgan's law, this is p and (q or ~q), which is p.