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Consider the open sentence P(x,y):x 2

+y 3
≤3, where x and y are real numbers. Establish the truth value of the implication: (∀x)(∃y)P(x,y)⇒(∃y)(∀x)P(x,y)

User Mbroshi
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Answer:

Explanation:

To establish the truth value of the implication (∀x)(∃y)P(x,y) ⇒ (∃y)(∀x)P(x,y), we need to examine both the antecedent and the consequent.

The antecedent (∀x)(∃y)P(x,y) states that for all values of x, there exists at least one value of y such that P(x,y) is true. In other words, it claims that every x-coordinate has at least one y-coordinate that satisfies the equation x^2 + y^3 ≤ 3.

The consequent (∃y)(∀x)P(x,y) states that there exists at least one value of y such that for all values of x, P(x,y) is true. In other words, it claims that there is at least one y-coordinate that satisfies the equation x^2 + y^3 ≤ 3 for every x-coordinate.

To determine the truth value of the implication, we need to evaluate whether the truth of the antecedent guarantees the truth of the consequent.

Now, let's analyze the given equation x^2 + y^3 ≤ 3:

Since x^2 is always non-negative for real numbers, in order for the equation to hold true, the term y^3 must be less than or equal to 3. This means that y must be between -∛3 and ∛3.

Considering this, we can conclude that for any value of x, we can always find at least one value of y that satisfies the equation x^2 + y^3 ≤ 3. Therefore, the antecedent (∀x)(∃y)P(x,y) is true.

Since the antecedent is true, it implies that the consequent (∃y)(∀x)P(x,y) must also be true. The implication is valid in this case.

Therefore, the truth value of the implication (∀x)(∃y)P(x,y) ⇒ (∃y)(∀x)P(x,y) is true.

User Tomaltach
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