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∃x (S(x) ∧ P(x)) ⇔ ¬∀x (S(x) → ¬P(x))

User HTN
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To determine the validity of the given statement, let's break it down and analyze it step by step using logical equivalences.

1) ∃x (S(x) ∧ P(x)) ⇔ ¬∀x (S(x) → ¬P(x))

First, let's simplify the right side of the equivalence using the negation of the implication:

2) ∃x (S(x) ∧ P(x)) ⇔ ¬∀x (¬S(x) ∨ ¬¬P(x))

Next, applying De Morgan's Law to the negation of the universal quantifier:

3) ∃x (S(x) ∧ P(x)) ⇔ ¬(∀x ¬S(x) ∨ ∀x ¬¬P(x))

Since the quantifiers are over the same variable, we can rewrite it as:

4) ∃x (S(x) ∧ P(x)) ⇔ ¬(∀x ¬S(x) ∨ ∃x P(x))

Now, using the negation of the disjunction:

5) ∃x (S(x) ∧ P(x)) ⇔ ¬(∀x ¬S(x)) ∧ ¬(∃x P(x))

Applying the negation to the universal quantifier:

6) ∃x (S(x) ∧ P(x)) ⇔ ∃x S(x) ∧ ¬(∃x P(x))

Finally, using De Morgan's Law again:

7) ∃x (S(x) ∧ P(x)) ⇔ ∃x S(x) ∧ ∀x ¬P(x)

The statement is now simplified. From this, we can observe that the left side of the equivalence states that there exists an x such that S(x) is true and P(x) is true. Meanwhile, the right side of the equivalence states that there exists an x such that S(x) is true and for all x, P(x) is false.

Therefore, the original statement is not valid.
User Cintra
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