Final answer:
The logical argument given, (∀y)(∃x)Q(x,y)→(∃x)(∀y)Q(x,y), is not valid. This is shown through a counterexample involving 'knowing' in a domain of people, where it's possible for everyone to be known by someone without a single person knowing everyone.
Step-by-step explanation:
The issue presented is related to logic and involves understanding whether the formula (∀y)(∃x)Q(x,y)→(∃x)(∀y)Q(x,y) is valid. To determine if an argument is valid, we look for a deductive inference where if the premises are true, then the conclusion must also be true automatically. Validity does not concern the actual truth of the premises or the conclusion, but rather, if the logical form dictates that the conclusion must follow from the premises.
For a consequence relation like 'if P, then Q', if P is sufficient for Q and P is proven to be true, Q must also be true. This type of reasoning is called modus ponens. Meanwhile, if Q is a necessary condition for P and Q is false, then P is also false, which is called modus tollens.
To provide a counterexample that shows the above formula is not valid, consider an interpretation with the domain of all people, and let Q(x, y) stand for "x knows y." It's quite possible that for every person y, there is some person x who knows them (meaning the premise (∀y)(∃x)Q(x,y) is true). However, it's not necessarily true that there exists a person x who knows every person y (which would make the conclusion (∃x)(∀y)Q(x,y) true). Therefore, the premise could be true without the conclusion being true, proving the argument is invalid.