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Recent questions tagged quantifiers

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8
UGC NET CSE | July 2018 | Part 2 | Question: 85
The equivalence of $\neg \: \exists \: x \: Q \: (x)$ is $\exists \: x \: \neg \: Q \: (x)$ $\forall \: x \: \neg \: Q \: (x)$ $\neg \: \exists \: x \: \neg \: Q \: (x)$ $\forall \: x \: Q \: (x)$
The equivalence of $\neg \: \exists \: x \: Q \: (x)$ is$\exists \: x \: \neg \: Q \: (x)$$\forall \: x \: \neg \: Q \: (x)$$\neg \: \exists \: x \: \neg \: Q \: (x)$$\f...
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13
Universal Quantifier (Basics)
Which one of the expression of universal quantifier is ambiguous? For all For every all of for each for any for arbitrary
Which one of the expression of universal quantifier is ambiguous?For allFor everyall offor eachfor anyfor arbitrary
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1 votes
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14
First Order Logic
A = ∃x (P(x) ^ Q(x)). B = ∃x P(x) ^ ∃x Q(x). Which is correct? a) A => B b) B => A c) A <=> B d) None of These Please Explain.
A = ∃x (P(x) ^ Q(x)).B = ∃x P(x) ^ ∃x Q(x).Which is correct?a) A = Bb) B = Ac) A <= Bd) None of ThesePlease Explain.
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15
First Order Logic
A = ∃x(P(x)^Q(x)) B = ∃x P(x) ^ ∃x Q(x), which is correct? a) A <=> B b) A => B c) B => A d) None of These Please Explain.
A = ∃x(P(x)^Q(x))B = ∃x P(x) ^ ∃x Q(x), which is correct?a) A <= Bb) A = Bc) B = Ad) None of ThesePlease Explain.
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16
Rosen_Discrete_Mathematics_and_Its_Applications_7th_Edition/Pg.56/Section 1.4 :predicates and quantifiers/Q.43 and Q.44
43. Determine whether ∀x(P(x)→Q(x)) and ∀xP (x)→∀xQ(x) are logically equivalent. Justify your answer.44. Determine whether ∀x(P(x)↔Q(x)) and ∀ x P (x) ↔...
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1 answer
20
Quantifiers
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3 votes
2 answers
21
Logic
Everyone has exactly one best friend Are all three below same? Let B(x, y) to be the statement “y is the best friend of x" $ ∀x∃y(B(x, y) ∧ ∀z((z = y)→¬B(x, z))) $ $ ∀x ∃!y (B(x, y) $ $ ∀x ∃y (B(x, y) ∧ ∀z (B(x, z) → (y = z))) $
Everyone has exactly one best friendAre all three below same?Let B(x, y) to be the statement “y is the best friend of x"$ ∀x∃y(B(x, y) ∧ ∀z((z = y)→¬B(x, z))...
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4 votes
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22
Multiplicative inverse
Every real number except zero has a multiplicative inverse Are both the statements same? $ ∀x((x != 0) → ∃y(xy = 1)) $ $ ∀x ∃y ((x != 0) → (xy = 1)) $
Every real number except zero has a multiplicative inverseAre both the statements same?$ ∀x((x != 0) → ∃y(xy = 1)) $$ ∀x ∃y ((x != 0) → (xy = 1)) $
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