Final answer:
The wff (∃x)[R(x)⋁S(x)]→(∃x)R(x)⋁(∃x)S(x) is proven valid through a disjunctive syllogism, and it is not possible to provide an interpretation in which it is false due to the inherent validity of this logical form.
Step-by-step explanation:
Proof of Validity
We are tasked to prove the validity of the well-formed formula (wff): ∃x)[R(x)⋁S(x)]→(∃x)R(x)⋁(∃x)S(x), or provide an interpretation where it is false. This logical statement uses quantifiers, disjunction, and implication. Let's analyze the components of this argument.
The initial part of the wff (∃x)[R(x)⋁S(x)] suggests that there exists an 'x' for which 'R(x)' or 'S(x)' is true. If this is the case, then logically it follows that either there exists an 'x' for which 'R(x)' is true, or there exists an 'x' for which 'S(x)' is true, which is what the conclusion states.
The structure of this argument resembles a disjunctive syllogism, which is a valid deductive inference. This means if there exists at least one instance where 'R(x)' or 'S(x)' is true, the conclusion that there exists an 'x' where 'R(x)' is true, or there exists an 'x' where 'S(x)' is true, must also be true. Therefore, the initial wff is indeed a valid argument by modus ponens.
Since we are determining the truth analysis of a wff, we are essentially engaging in a form of logical reasoning where we assess whether the premises guarantee the truth of the conclusion. The modus ponens inference used in our proof makes the wff valid, and trying to find a counterexample would not disprove it because the logical form itself is valid.
Finding a false interpretation for this valid argument is not possible because, by its form, if the premise is true, the conclusion necessarily follows. Any other claimed counterexample would stem from misunderstanding the logical structure and possibly creating a strawman fallacy. In summary, the given wff is a valid argument that demonstrates the correctness of the original conditional statement given the truth of its premise.