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this is vacuously true for all elements, x, in the empty set, P(x) is true there is no x to falsify the claim of P(x) so you can conclude the statement is true
I've been doing some research and I found that this is a shorthand for \[x\in \emptyset \Rightarrow P(x) \] Which is TRUE, but I don't see the relationship.
But that is like a convention because there is no x to make TRUE the statement.
doesn't matter the only thing that causes a condition to be false is if "True implies False" in this case, the conditional statement is always false, so it can imply anything, and the entire statement will always evaluate to true
in other words, the only thing that causes a conditional statement to be false is if the conditional is true AND the consequent is false
I understand what you say about the conditional statement, but what I'm not sure is how you can go from the universal quantifier to a conditional statement.
\[x\in \emptyset \Rightarrow P(x)\] totally makes sense to me. What I can't see is how is this equivalent to the universal quantification.
Sorry, can anyone just tell me what \(\rm P(x)\) stands for?
I do know that conditional is not equivalent to \(\rm if\cdots then\) statements.
Oh, I get it.
\[P(x)\] is a predicate with a parameter x. Once you know the value of the parameter x you can say it becomes a proposition and you can say if its TRUE or FALSE. For example \[x > 0\]
if you give x the value -1 you can say that the proposition is FALSE.
\[\forall x \in \emptyset :\rm P(x)\]This is true, means that \(\rm P(x)\) is true for all \(x\) in empty set. If a conditional is true, then: * The first value is false. * If not false, then the second value must be true. In other words \(1\implies 0\) is false.
So \(x \in \emptyset \implies P(x)\) is true. If \(x \not \in \emptyset\), then the first part of the conditional is false. So the statement is true.
If \(x \in \emptyset\), we know that \(P(x)\) is true so the conditional is true.
So the statement is always true, and it is given in your original statement that it is true.
Two statements having the same truth tables are equivalent.
what i was trying to say earlier was that a universal quantifier states a case for any x in the empty set so an equivalent statement is exactly as you have written you can state the case for any x in the empty set as \[(x \in \emptyset) \rightarrow P(x)\]
Then can I turn every universal quantification into an implication?
the only difference is in the notation. \[\forall x \in \emptyset, P(x)\] just says "for any x in the empty set, P(x) is true" whereas \[(x \in \emptyset) \rightarrow P(x)\] just says "if x is in the empty set, then P(x) is true"
yes you can
generally speaking \[\forall x \in X, P(x)\] is equivalent to \[(x \in X) \rightarrow P(x)\]