Give interpretations to prove that each of the following wffs is not valid:a. Ex)A(x) ^ (Ix)B(x) → Ex)[A(x) B(x)]b. (Vx)Ay)P(x, y)→Ex)(Vy)P(x. y)c. (Vx)[P(x)→ Q(x)] → [@x)P(x) → (x)Q(x)]d. (Vx)[A(x)]' → [(Vx)4(x)]'

a. (∃x)A(x) ∧ (∃x)B(x) → (∃x)[A(x) ∧ B(x)] This wff is invalid. Since, the existential quantifier…

Answered: Give interpretations to prove that each…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Give English language translations of the following wffs if M(x): x is a man W(x): x is a woman i: Ivan p: Peter W(x, y): x works for y a. (Ax)(W(x) ^ (Vy)(M(y) → [W(x, y)1). b. (Vx)[M(x) → 3y)(W(y) ^ W(x, y))] c. (Vx)[M(x) → (Vy)(W(x, y) → W(y))] d. (Vx)(Vy)(M(x) A W(y, x) → W(y)) e. W(i, p) ^ (Vx)[W(p, x) → (W(x))') f. (Vx)[W(x, i) → (W(x))']
Given: ∀x [p(x) → q(x)]∀x[p(x) → (∀y w(y))]∀x ∀y [(q(x) ∧ w(y)) → s(x)] p(MARY) Show s(MARY) using resolution.
Prove that each of the following is a valid argument: (vx)P(x) A (3x)Q(x) → (3x)[P(x) ^ Q(x)] (Vx)(vy)Q(x, y) → (vy)(x)Q(x, y)
6. Find g •f (a) f:Ž– N, f(n) = n² + 1; g:N → Q, g(n) = g: (0, 1) → (0, 1), g(x) =1– x. (b) f:R → (0, 1), f(x)= 1/(x² + 1); (c) f:Q – {2} → Q*, f(x) = 1/(x – 2); (d) f:R → [1,00), f(x)= x² + 1; g: Q* → Q*, g(x) = 1/x. g:[1, 00) → [0, 0) g(x)= /r – 1. + [0, 00) g(x) = Vx – 1. (e) f:Q – {10/3} → Q – {3}, f(x) = 3x – 7; 2r /(т — 3). g: Q – {3} → Q – {2}, g(x) =
Given: ∀x [p(x) → q(x)]∀x [p(x) → (∀yw(y))]∀x∀y [(q(x) ∧ w(y)) → s(x)] p(MARY) Show s(MARY) using resolution.
i. Translate Vx[C(x) V3y(C(v) A F(xy)] into English, where C(x): x has a computer F(x y): x and y are friends The domain for both x and y is all students in your school. ii. Translate 3xvyz [[F(xv)^F(xz)^(y =6 z)) -→¬F(y,z)] where; F(a,b): a and b are friends. The domain of x, y and z consists of all students in your school.
Interpret this in plain english Φ = (a ⇒ b) ^ (c ⇒ b’) ^ (a’⊕ c)
Question 6 Simplify the following Boolean expressions using K-Map a- F(w,x,y,z)= (1,3,4,5,6,7,9,11,13,15) b- B'D'(A'C'+C)+AC'D'+BD(A'C+C')
Use resolution to show Vx Vy[p(x, y) → ¬p(y, x)] F ¬p(x, x)
1/3 Let W = [(a, b, c, d)| 5a - b=0 ) be a subspace of R'. Then a basis for W is O {(1,5,0,0)} O {(1,5,0,0),(0,0,0,1)} O {(1,5,0,0),(0,0,1,0),(0,0,0,1)} {(5,1,0,0),(0,0,1,0),(0,0,0,1)}
Let M be the PDA defined by Q = {q, q0 , q1 , q2}, Σ = {a,b}, Γ = {a}, F := {q , q1}.δ(q0 , a , Z0) = {(q, Z0 ) } δ(q , a, Z0) = {(q , aZ0)} δ(q , a, a) = {(q , aa)} δ(q , b , a) = {(q1 ,e)} δ(q1 , b , a) = {(q1 ,e)} δ(q1 , b, Z0 ) = {(q2 , e)} Describe the CFG and the language accepted by M
either prove that the wff is a valid argument or give an interpretation in which it is false. [(∀x)P(x) → (∀x)Q(x)] → (∀x)[P(x) → Q(x)]

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Give interpretations to prove that each of the following wffs is not valid: a. Ex)A(x) ^ (Ix)B(x) → Ex)[A(x) B(x)] b. (Vx)Ay)P(x, y)→Ex)(Vy)P(x. y) c. (Vx)[P(x)→ Q(x)] → [@x)P(x) → (x)Q(x)] d. (Vx)[A(x)]' → [(Vx)4(x)]'