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Archangel Macsika

For each proposition below, decide whether it is true or false and give a brief explanation. Assume the universe (domain of variables) to be Z, the set of integers. (1) ((x = 5) ∧ (y = 1)) → ((x > 10) ∨ (y > 0)) (2) ∀x((x < 0) ∨ (x^2 ≥ x)) (3) ¬(∃xP(x)) ↔ (∀x¬P(x)) for all predicates P(x)

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For each proposition below, let us decide whether it is true or false. Assume thedomain of variables to be \Z, the set of integers.


(1) ((x = 5) ∧ (y = 1)) → ((x > 10) ∨ (y > 0))

The conjunction (x = 5) ∧ (y = 1) is true if and only if x=5 and y=1. In this case the disjunction (x > 10) ∨ (y > 0)=(5 > 10) ∨ (1 > 0) is true, and hence the implication ((x = 5) ∧ (y = 1)) → ((x > 10) ∨ (y > 0)) is also true. In other cases the conjunction (x = 5) ∧ (y = 1) is false, and hence the implication ((x = 5) ∧ (y = 1)) → ((x > 10) ∨ (y > 0)) is true. Therefore, for all integers x and y the statement ((x = 5) ∧ (y = 1)) → ((x > 10) ∨ (y > 0)) is true.

Answer: true


(2) ∀x((x < 0) ∨ (x^2 ≥ x))

Since for each integer x the statement x^2\ge x is true, we conclude the disjunction (x < 0) ∨ (x^2 ≥ x) is also true, and hence the statement ∀x((x < 0) ∨ (x^2 ≥ x)) is true.

Answer: true


(3) ¬(∃xP(x)) ↔ (∀x¬P(x)) for all predicates P(x)

Since ¬(∃xP(x)) \equiv (∀x¬P(x)), the statements ¬(∃xP(x)) and (∀x¬P(x)) have the same values, and hence the equivalence ¬(∃xP(x)) ↔ (∀x¬P(x)) is true.

Answer: true


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