Question

Identify the error or errors in this argument that supposedly shows that if ∀x(p(x)∨q(x)) is true then ∀x p(x) ∨ ∀x q(x)) is true?

Answer 1

The error in the argument that supposedly shows if
`\( \forall x(p(x) \lor q(x)) \)` is true, then
`\( \forall x p(x) \lor \forall x q(x) \)` is a misunderstanding of the distribution of quantifiers. The correct form
`\( \forall x(p(x) \lor q(x)) \)` does not necessarily imply
`\( \forall x p(x) \lor \forall x q(x) \)`.

Step-by-step explanation:

The error in the argument lies in misinterpreting the distribution of the universal quantifier
`(\( \forall \))`. In formal logic,
`\( \forall x(p(x) \lor q(x)) \)` means "for every x, either p(x) or q(x) (or both) is true." However, it does not imply that "for every x, p(x) is true or for every x, q(x) is true."

To illustrate this, consider a scenario where p(x) is "x is even" and q(x) is "x is odd." The statement
`\( \forall x(p(x) \lor q(x)) \)` would be true because for every x, either p(x) (x is even) or q(x) (x is odd) is true. However,
`\( \forall x p(x) \lor \forall x q(x) \)` would be false because not every x is even, and not every x is odd.

In summary, the error arises from an incorrect inference about the distribution of the universal quantifier, leading to a faulty conclusion about the relationship between
`\( \forall x(p(x) \lor q(x)) \)` and
`\( \forall x p(x) \lor \forall x q(x) \)`.