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Let $a(x, y), b(x, y,)$ and $c(x, y)$ be three statements with variables $x$ and $y$ chosen from some universe. Consider the following statement:

$\qquad(\exists x)(\forall y)[(a(x, y) \wedge b(x, y)) \wedge \neg c(x, y)]$

Which one of the following is its equivalent?

  1. $(\forall x)(\exists y)[(a(x, y) \vee b(x, y)) \to c(x, y)]$
  2. $(\exists x)(\forall y)[(a(x, y) \vee b(x, y)) \wedge\neg c(x, y)]$
  3. $\neg (\forall x)(\exists y)[(a(x, y) \wedge b(x, y)) \to c(x, y)]$
  4. $\neg (\forall x)(\exists y)[(a(x, y) \vee b(x, y)) \to c(x, y)]$
in Mathematical Logic edited by
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1 Answer

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Best answer

$(\exists x)(\forall y)[(a(x, y) \wedge b(x, y)) \wedge ¬c(x, y)]$

$\quad \equiv ¬(\forall x)¬(\forall y)[(a(x, y) \wedge b(x, y)) \wedge ¬c(x, y)]$
$\quad (\because (\exists x) F(x) = \neg \forall x \neg F(x))$

$\quad \equiv ¬(\forall x)(\exists y)¬[(a(x, y) \wedge b(x, y)) \wedge ¬c(x, y)]$
$\quad (\because (\forall x) F(x) = \neg \exists x \neg F(x)), \neg \neg F(x) = F(x))$

$\quad \equiv ¬(\forall x)(\exists y)[¬(a(x, y) \wedge b(x, y)) \vee c(x, y)]$

$\quad \equiv ¬(\forall x)(\exists y)[(a(x, y) \wedge b(x, y)) → c(x, y)]$

(C) choice.

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9 Comments

Please explain Sir,how the first step is derived
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That is Demorgan's law-

$$\exists x f(x) = \neg \forall x \neg f(x)$$

Also

$$\forall x f(x) = \neg \exists x \neg f(x)$$

In English we can say that if f(x) is true for all x, then there does not exist any x for which f(x) is not false. Similarly, if there exist an x for which f(x) is true, then it is not the case that f(x) is false for all x.
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how 2nd step is calculated?

plz explain sir
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it is the same Demorgan's law. See now..
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Nice Explanation Arjun sir
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edited by
but @Arjun sir

Demorgan law is

¬∀xP(x) ≡ ∃x¬P(x)

¬∃xP(x) ≡ ∀x¬P(x)

then why did u write it as

∃xf(x)=¬∀x¬f(x)

please clear my doubt sir
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@Arnab , ¬∀x¬f(x) = ∃x¬¬f(x) = ∃xf(x)(because by demorgan's law we are getting this)
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srestha

what is now not in syllabus??

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edited by

@akash

I thought it is logic programming

like this https://gateoverflow.in/958/gate2003-71

sorry misunderstood

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Answer:

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