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Here are some statements that I had to translate into symbolic form, then negate and express the negation as a positive statement. In each case the assumed domain is given in parentheses.

a. Everyone loves somebody. (All people)

Let:
$L(x,\;y)$ mean $x$ loves $y$.

Symbolic form: $\forall x \exists y [L(x,\;y)]$

Negation: $\exists x \forall y\;[\lnot L(x,\;y)]$

Negated statement in english: Someone doesn't love everyone.

b. Nobody loves everybody. (All people)

Let:
$L(x,\;y)$ mean $x$ loves $y$.

Symbolic form: $\lnot\exists x \forall y [L(x,\;y)]$

Negation: $\exists x \exists y [\lnot L(x,\;y)]$

Negated statement in english:
"Someone doesn't love someone"

or

"At least one person doesn't love someone"

c. If a man comes, all the women will leave. (All people)

Let:
$M(x)$ mean $x$ is a man.
$C(x)$ mean $x$ comes.
$W(x)$ mean $x$ is a woman.
$L(x)$ mean $x$ is a leaves.

Symbolic form: $\exists x {\big[}M(x) \land C(x){\big]} \Rightarrow \forall x{\big[}W(x) \Rightarrow L(x){\big]}$

Negation:
$\exists x {\big[}M(x) \land C(x){\big]} \nRightarrow \forall x{\big[}W(x) \Rightarrow L(x){\big]}$
$\exists x {\big[}M(x) \land C(x){\big]} \land \exists x{\big[}W(x) \land \lnot L(x){\big]}$

Negated statement in english: At least one man will come at least one woman will stay.

d. Not all precious stones are beautiful. (All stones)

Let:
$P(x)$ mean $x$ is a precious stone.
$B(x)$ mean $x$ is beautiful.

Symbolic form: $\exists x\;[P(x) \land \lnot B(x)]$

Negation
$\exists x\;[P(x) \land \lnot B(x)]$ is equivalent to $\exists x\;[P(x) \nRightarrow B(x)]$ where $(a \land \lnot b) \Leftrightarrow (a \nRightarrow b)$
So the negated version is: $\forall x\;[P(x) \Rightarrow B(x)]$

Negated statement in english: All precious stones are beautiful.

e. Nobody loves me. (All people)

Let:
$L(x,\;y)$ mean $x$ loves $y$.
$m$ mean $me$.

Symbolic form: $\lnot \exists x [L(x,\;m)]$

Negation: $\exists x [\lnot L(x,\;m)]$

Negated statement in english: Somebody doesn't love me.

f. At least one American snake is poisonous. (All snakes)

Let:
$A(x)$ mean $x$ is american.
$P(x)$ mean $x$ is poisonous.

Symbolic form: $\exists x [A(x) \land P(x)]$

Negation:
$\forall x\;[\lnot A(x) \lor\lnot P(x)]$
$\forall x\;[A(x) \Rightarrow\lnot P(x)]$; where $(\lnot a \lor b) \Leftrightarrow (a \Rightarrow b)$.

Negated statement in english: No american snake is poisonous.

g. At least one American snake is poisonous. (All animals)

Let:
$S(x)$ mean $x$ is a snake.
$A(x)$ mean $x$ is american.
$P(x)$ mean $x$ is poisonous.

Symbolic form: $\exists x [S(x) \land A(x) \land P(x)]{\big]}$

Negation
$\forall x\;[\lnot S(x) \lor \lnot A(x) \lor \lnot P(x)]$
Grouping together $A(x)$ and $S(x)$:
$\forall x\;{\big[}\lnot [\lnot S(x) \lor \lnot A(x)] \Rightarrow \lnot P(x){\big]}$; where $(\lnot a \lor b) \Leftrightarrow (a \Rightarrow b)$
$\forall x\;{\big[} [ S(x) \land A(x)] \Rightarrow \lnot P(x){\big]}$

Negated statement in english: No american snake is poisonous.

ryang
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Carlos
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2 Answers2

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a. Checks out symbolically, but I would recommend the wording "somebody doesn't love anyone."

b. Your negation is a bit suspicious here, you can just cancel out the leading negation. You "distribute" the negation inwards as if you applied DeMorgan's without actually using DeMorgan's. (this of course trickles down to the written negation)

c. Checks out from what I can tell, the wording is a bit awkward but I can't think of a better alternative.

d. Good

e. Again, for the negation you can just get rid of the leading negation.

f. Good

g. Good, except for a minor typo where you put $y$ where an $x$ should be

Overall, I think you've done well but there are a few errors which I think could be avoided if you write out more of your intermediate steps and focus on the logic which moves you from one step to the next. Once you've become more accustomed to this then I think it'll feel a bit more natural and you can skip steps more safely.

Also, I recommend using your written statements to check your answers: starting from the first statement and not considering the symbolic logic, how would you negate the statement if you were speaking? If it disagrees with the answer you end up with, you should double-check your symbolic argument.

Hope this helps!

  • Hi @Stephen Donovan. Thanks for pointing out how my use of $\lnot$ was dubious. In fact, it was wrong, I was using it to negate $\exists$ to turn it into does not exist. I think the correct symbolic form is: $$\lnot{\big[}\exists x;\forall y;[L(x,;y)]{\big]}$$ $$\forall x;\exists y;[\lnot L(x,;y)]$$

    And the negative form is: $$\exists x;\forall y;[ L(x,;y)]$$

    Which means Somebody loves everybody.

    I actually found the solution to the same exercise in the book Discrete mathematics and its applications by Rosen (7th ed).

    Cheers.

    – Carlos Nov 07 '21 at 07:07
  • I think e is wrong. Nobody loves me should be $\forall x;[\lnot L(x,;me)]$. The negated form should be $\exists x;[L(x,;me)]$, which reads Somebody loves me. – Carlos Nov 07 '21 at 07:41
  • @Carlos Those both look like suitable corrections for b and e. To clarify, I agreed with your initial translations in both cases, it's fine to have the $\neg$ out in front the way you did, what I disagreed with was how you negated them. In such cases you can just use the double negative rule to cancel out the negative in front. So for your original translation for b, notice that removing the leading $\neg$ gives you $\exists x \forall y [L(x,y)],$ which is the same as your new answer. – Stephen Donovan Nov 08 '21 at 14:02
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a. Everyone loves somebody. (All people)

Symbolic form: $\forall x \exists y [L(x,\;y)]$

Negation: $\exists x \forall y\;[\lnot L(x,\;y)]$

Negated statement in english: Someone doesn't love everyone.

Correction: "Someone doesn't love anybody". (Your symbolic negation is correct though.)

If everybody loves Raymond and Raymond loves only himself, then your suggested statement is True while my correction statement is False.

An orthogonal point worth raising: the given sentence is technically ambiguous, because it could have been alternatively mechanically translated as $\exists x \forall y [L(y,\;x)]$ due to the last word being a hanging quantifier.

b. Nobody loves everybody. (All people)

Symbolic form: $\lnot\exists x \forall y [L(x,\;y)]$

Negation: $\exists x \exists y [\lnot L(x,\;y)]$

Negated statement in english: "Someone doesn't love someone" or "At least one person doesn't love someone"

The negation (just drop that negation symbol!) ought to be $$\exists x \forall y [L(x,\;y)]$$ "Somebody loves everybody" instead.

c. If a man comes, all the women will leave. (All people)

Negated statement in english: At least one man will come at least one woman will stay.

No need to change the tense: "A man comes and some woman will stay".

d. Not all precious stones are beautiful. (All stones)

Symbolic form: $\exists x\;[P(x) \land \lnot B(x)]$

Isn't the symbolic form merely $$\lnot \forall x\;[P(x) \Rightarrow B(x)]?$$

Negation $\forall x\;[P(x) \Rightarrow B(x)]$

Negated statement in english: All precious stones are beautiful.

Correct.

e. Nobody loves me. (All people)

Symbolic form: $\lnot \exists x [L(x,\;m)]$

Negation: $\exists x [\lnot L(x,\;m)]$

Negated statement in english: Somebody doesn't love me.

The negation (just drop that negation symbol!) ought to be $$\exists x [L(x,\;m)]$$ "Somebody loves me".


f and g are both correct.


Addendum

OP (emphasis mine): The problem I was pointing out is the order of precedence that ¬ has. I used it as to negate just the first quantifier, whereas it seems that you take it to negate the whole expression, even if it's not made explicit by means of parentheses.

Sentences (e.g., ‘the rose is red’, ‘the number of marbles is $7$’, ‘$2x-7=0$’) can be negated. Expressions (e.g., ‘red’, ‘the number of marbles’, ‘$2x-7$’) and quantifiers (e.g., $∀y$) can neither be true nor false, so how can their truth value be flipped, in other words, how can they be negated?

Facts (inserting parentheses to demonstrate that order of precedence is irrelevant): \begin{align}(¬∀x)\:P(x)\;&\equiv\;∃x\:¬P(x)\\&\boldsymbol{\not\equiv}\;∃x\:P(x)\;;\\(¬∀x)∃y\:P(x,y)\;&\equiv\;∃x∀y\:¬P(x,y)\\&\boldsymbol{\not\equiv}\;∃x∃y\:P(x,y).\end{align}

After all, the negation of $$\text{not all xylophones are pink}$$ is $$\text{some xylophone is not pink},$$ rather than $$\text{some xylophone is pink}.$$

If this addendum has revised your understanding, then my above responses to Parts b & d, and my first comment below, will now make more sense.

ryang
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  • I think d is correct. I don't think doing $\lnot \forall$ or $\lnot \exists$ is correct. Furthermore, if you start with all precious stones are beautiful, $\forall x;[P(x) \Rightarrow B(x)]$, and then negate it, you have to turn the $\forall$ into a $\exists$ and the $\Rightarrow$ into $\nRightarrow$. Then you can consider that $(a \land \lnot b) \Leftrightarrow (a \nRightarrow b)$ . You end up with my initial symbolic form. – Carlos Nov 07 '21 at 07:36
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    @Carlos Your (d) translation AND negation are both correct, and I hadn't suggested otherwise. I was pointing out that "Not all precious stones are beautiful" merely and literally translates to " ¬∀x[P(x)⇒B(x)] " (the "∃x[P(x)∧¬B(x)]" that you suggest is not a direct translation, but nonetheless logically equivalent), and its subsequent negation obtained by merely dropping the negation symbol from this. – ryang Nov 07 '21 at 14:31
  • The problem I was pointing out is the order of precedence that $\lnot$ has. I used it as to negate just the first quantifier, whereas it seems that you take it to negate the whole expression, even if it's not made explicit by means of parentheses. – Carlos Nov 10 '21 at 22:23
  • @Carlos Too long for a comment; added an addendum instead. – ryang Nov 11 '21 at 04:24
  • I think your initial proposed answer to d was correct. I would just add the parentheses for clarity's sake. Statement: Not all precious stones are beautiful. (All stones). Symbolic form: $\lnot{\big [}\forall x;[P(x) \Rightarrow B(x)]{\big ]}$. Negated form: $\forall x;[P(x) \Rightarrow B(x)]$. In english: All precious stones are beautiful. – Carlos Nov 13 '21 at 07:29
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    @Carlos My answer has never changed, and has been correct all along: now you're merely duplicating it into your latest comment, and adding an unnecessary parenthesis $(¬∀x:P(x)$ is *never* ever understood as $∃:P(x),$ because it does not make sense to speak/think of negating a quantifier!) In any case, it's good that your misunderstanding has finally cleared up! – ryang Nov 13 '21 at 07:52
  • Thanks a lot for the comments. I up-voted both answers long ago because they both were very helpful.
    I think you're example of the Xilophones is wrong though.
    Statement: not all xylophones are pink.
    Negation: all xylophones are pink.
    – Carlos Nov 15 '21 at 10:33
  • @Carlos You keep going around in circles. Your latest comment is correct, and so is my Xylophone example. The point of that example was to explain to you that negating a quantified statement is not just about flipping the quantifier symbol. To be extremely clear: to negate "not all xylophones are pink", we can just drop the "not" (as you have just done), or we can write "some xylophone is not pink" (as in my explanation), but it would be wrong to write "some xylophone is pink." – ryang Nov 15 '21 at 10:53