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This might be very easy for anyone who understand the concepts, but I think it highlights how little my understanding is. I have $\neg \exists x(P(x)\vee Q(x) \to R(x) \vee S(x))$.

I would like to bring the negation inwards, so I did $\forall x \neg (P(x)\vee Q(x) \to R(x) \vee S(x))$, and here I am stuck.

Should I see it as $\forall x \neg (P(x)\vee B)$, and proceed to negate the disjunction, and then to negate the components of the disjunction, or is the order different? My teacher never published an example in which the parentheses are ambiguous, like this one. Usually clauses are grouped in couples, but not this time, and I am a bit lost.

Thanks for all the help!

Foxtrot_Romeo
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1 Answers1

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Since your expression is lacking parentheses and/or brackets, I will assume the normal rules of operator precedence apply: $(1) \neg (2) \wedge (3) \vee (4) \to (5) \leftrightarrow$. In order to make the precedence (and computation) more clear, I add some brackets to the expression and then move negation inwards.

$\neg \exists x\Big ( \Big[ P(x)\vee Q(x) \Big] \to \Big[ R(x) \vee S(x) \Big]\Big)$

$\forall x \neg \Big ( \Big[ P(x)\vee Q(x) \Big] \to \Big[ R(x) \vee S(x) \Big]\Big)$ DeMorgan's law for quantifiers

$\forall x \neg \Big ( \neg \Big[ P(x)\vee Q(x) \Big] \vee \Big[ R(x) \vee S(x) \Big]\Big)$ implication law

$\forall x \Big ( \neg \neg \Big[ P(x)\vee Q(x) \Big] \wedge \neg \Big[ R(x) \vee S(x) \Big]\Big)$ DeMorgan's law

$\forall x \Big ( \Big[ P(x)\vee Q(x) \Big] \wedge \neg \Big[ R(x) \vee S(x) \Big]\Big)$ double negation law

$\forall x \Big ( \Big[ P(x)\vee Q(x) \Big] \wedge \Big[ \neg R(x) \wedge \neg S(x) \Big]\Big)$ DeMorgan's law

$\forall x \Big ( \Big[ P(x)\vee Q(x) \Big] \wedge \neg R(x) \wedge \neg S(x) \Big)$ brackets can be dropped due to associative law

RyRy the Fly Guy
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