2
$\begingroup$

Negation of "Either X is true, or Y is true, but not both"

My attempt:

If seems that let X be true and Y be true, not X for X is false and not Y for Y is false. In order for the above statement to be True, we need:

The negation of both X and Y to be true: negate(X and Y) -> not X or not Y

For "Either X is true, or Y is true, but not both" is equivalent to below:

((X or Y) and (not X or not Y))

The negation of the above (negation turns "and" into "or", and turns "or" into "and"):

((not X and not Y) or (X and Y))

Is this logical or? I am pretty lost...

$\endgroup$
  • 4
    $\begingroup$ This would be logical biconditional, i.e., the negation is true if and only if $X$ and $Y$ have the same truth value. But your work is correct. An easier way to try to visualize this might be with a truth table. $\endgroup$ – Hayden Jun 3 at 22:06
  • $\begingroup$ This proof tree might help. $\endgroup$ – Shaun Jun 3 at 22:24
  • $\begingroup$ Why not just X=Y? $\endgroup$ – n0rd Jun 4 at 22:09
7
$\begingroup$

A basic approach is to see there only are 4 possibilities:

  1. $X$ and $Y$
  2. $X$ and not $Y$
  3. not $X$ and $Y$
  4. not $X$ and not $Y$

Your statement was (2) or (3), so its negation is (1) or (4).

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

I give a solution by formal calculation: translate the phrase into logical symbols, we have $\text{negation}\rightarrow\neg$, $\text{either X or Y is true}\rightarrow X\lor Y$, $\text{but not both}\rightarrow\land\neg(X\land Y)$. Then

$\neg((X\lor Y)\land\neg(X\land Y)=\neg(X\lor Y)\lor(\neg\neg(X\land Y))=(\neg X\land\neg Y)\lor(X\land Y)$

The final result is identical to yours.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Either $X$ is true, or $Y$ but not both is

($X$ OR $Y$) AND (NOT [$X$ AND $Y$])

Now the negation of $A$ AND $B$ is:

(not A) OR (not B).

So the negation is:

[NOT (X OR Y)] OR (NOT(NOT([X AND Y]))

And NOT(NOT(A)) is .. A so

[NOT (X OR Y)] OR [X AND Y]

And the negation of A OR B is: (NOT A) AND (NOT B).

So

(NOT X AND NOT Y) OR (X AND Y)

So the negation is

Either both X and Y are true, or both X and Y are false.

Maybe that is what it intuitively what you would have thought.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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