State division and remainder algorithm. Suppose that the domain of the propositional function P(x) consists of the integer 0, 1, 2, 3 and 4. Write out each of the following propositions using disjunctions, conjunctions and negations.

  1. ∃x P(x)
  2. ∀x P(x)
  3. ∃x ¬P(x)
  4. ∀x ¬P(x)
  5. ¬∃x P(x)
  6. ¬∀x P(x)

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Division and Remainder Algorithm:

Let a be an integer and ‘d’ be positive integer then there exists two unique integer q and r with 0 < r < d such that a = dq + r

where

a = dividend

d = divisior

q = quotient

r = remainder

i.e.

dividend = divisor * quotient + Remainder

For Example:

let a = 19, d = 3

3 ) 19 ( 6
   -18
  -----
     1
19 = 3 x 6  + 1

Second Part:

As we know that Universal Quantifier can be expressed as conjunctions and Existential Quantifier can be expressed as Disjunctions, given functions can be written as following:

a) ∃x P(x) = P(0) ∨ P(1) ∨ P(2) ∨ P(3) ∨ P(4)

b) ∀x P(x) = P(0) ∧ P(1) ∧ P(2) ∧ P(3) ∧ P(4)

c) ∃x ¬P(x) =  ¬P(0) ∨ ¬P(1) ∨ ¬P(2) ∨ ¬P(3) ∨ ¬P(4)

d) ∀x ¬P(x) = ¬P(0) ∧ ¬P(1) ∧ ¬P(2) ∧ ¬P(3) ∧ ¬P(4)

e) ¬∃x P(x) = ¬ (P(0) ∨ P(1) ∨ P(2) ∨ P(3) ∨ P(4))

f) ¬∀x P(x) = ¬ (P(0) ∧ P(1) ∧ P(2) ∧ P(3) ∧ P(4))

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