- E34
[016]As already commented in [00R], given \(A\) a set, and \(P(x)\) a logical proposition dependent from a free variable \(x\), we usally write
\[ ∀ x∈ A, P(x)\quad ,\quad ∃ x∈ A, P(x) \]however
\[ ∀ x∈ A, P(x)~ ~ \text{summarizes} ~ ~ ∀ x, (x ∈ A) ⇒ P(x)~ ~ , \]\[ ∃ x∈ A, P(x)~ ~ \text{summarizes} ~ ~ ∃ x, (x ∈ A) ∧ P(x)~ ~ ; \]where the ”extended” versions are well-formed formulas.
Using this extended version you can prove that the two propositions
\[ ¬(∀ x∈ A, P(x))~ ~ ,~ ~ ∃ x∈ A, (¬ P(x))~ ~ . \]are equivalent, in the sense that from one it is possible to prove the other (and vice versa). In the proof use only tautologies (listed in [00N]) and in particular the equivalence of the formula ”\(P⇒ Q\)” with ”\((¬ P)∨ Q\)” 1 , and finally the equivalence between ”\(¬ ∃ x, Q\)” and ”\(∀ x, ¬ Q\)” 2 .
Replacing \(P(x)\) with \(¬ P(x)\) and using the tautology of double negation finally results in
\[ ∀ x∈ A, (¬ P(x))~ ~ ,~ ~ ¬(∃ x∈ A, P(x))~ \]are equivalent.
Solution 1
EDB — 016
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English
Authors:
"Mennucci , Andrea C. G."
.
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