SOLUTION: Use the finite universe method to prove that the following argument is invalid: 1. (x)Ax⊃(∃x)Bx 2. (∃x)Ax / (∃x)Bx

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Question 1188876: Use the finite universe method to prove that the following argument is invalid:

1. (x)Ax⊃(∃x)Bx
2. (∃x)Ax / (∃x)Bx

Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Here's how to use the finite universe method to demonstrate the invalidity of the argument:
1. Define a Finite Universe:
Let the universe consist of just one object, which we'll call "a." So, our universe U = {a}.
2. Interpret the Premises and Conclusion:
Premise 1: (x)Ax ⊃ (∃x)Bx becomes (Aa ⊃ Ba). This means "If everything has property A, then there is something that has property B."
Premise 2: (∃x)Ax becomes Aa. This means "There is something that has property A" (In our universe, this means 'a' has property A).
Conclusion: (∃x)Bx becomes Ba. This means "There is something that has property B" (In our universe, this means 'a' has property B).
3. Assign Truth Values:
We need to assign truth values (True or False) to the properties A and B for the object 'a' in such a way that the premises are true, but the conclusion is false.
Let Aa be True. ('a' has property A).
Let Ba be False. ('a' does not have property B).
4. Evaluate the Premises:
Premise 1: (Aa ⊃ Ba) becomes (True ⊃ False), which is False.
Since premise 1 is false, the argument is invalid.
Explanation:
We have created a scenario where 'a' has property A (Aa is True) and 'a' does not have property B (Ba is False). This makes Premise 2 (∃x)Ax true. However, the conclusion (∃x)Bx is false.
The crucial point is Premise 1 (x)Ax ⊃ (∃x)Bx. This premise becomes false in our model. Since we could make the premises true while the conclusion is false, the argument is invalid.

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