Question

Which of the following logical statements is not valid?

• A. \( \forall x\, P(x) \Rightarrow \exists x\, \neg P(x) \)
• B. \( \forall x\, P(x) \Rightarrow \exists x\, P(x) \)
• C. \( \exists x\, P(x) \Rightarrow \forall x\, P(x) \)
• D. \( \exists x\, P(x) \Leftrightarrow \forall x\, P(x) \)

Hint:

Remember: Universal truth implies existence, but existence does not imply universal truth.

Answer 1

The Correct Option is A

Step 1: Understand the meaning of quantifiers.
\( \forall x\, P(x) \) means \( P(x) \) is true for every element.
\( \exists x\, P(x) \) means there exists at least one element for which \( P(x) \) is true.
\( \exists x\, \neg P(x) \) means there exists at least one element for which \( P(x) \) is false.
Step 2: Check option (A).
If \( \forall x\, P(x) \) is true, then \( P(x) \) is true for every element.
So it is impossible that \( \exists x\, \neg P(x) \) is true.
Thus the implication \[ \forall x\, P(x) \Rightarrow \exists x\, \neg P(x) \] is logically invalid.
Step 3: Check option (B).
If \( \forall x\, P(x) \) is true, then certainly at least one element satisfies \( P(x) \).
So this implication is valid.
Step 4: Check option (C).
If there exists one element satisfying \( P(x) \), it does not necessarily mean all elements satisfy it.
Thus this implication is not universally valid.
Step 5: Check option (D).
Existence of one element satisfying \( P(x) \) does not imply all satisfy it, so biconditional is not valid.
However, among the given options, the clearly invalid direct contradiction is option (A).