Question #264578

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)





Expert's answer

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


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

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

Answer: true


(2) x((x<0)(x2x))∀x((x < 0) ∨ (x^2 ≥ x))

Since for each integer xx the statement x2xx^2\ge x is true, we conclude the disjunction (x<0)(x2x)(x < 0) ∨ (x^2 ≥ x) is also true, and hence the statement x((x<0)(x2x))∀x((x < 0) ∨ (x^2 ≥ x)) is true.

Answer: true


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

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

Answer: true


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