Question

Can someone explain this in plain english, or spanish:
\[\forall x \in \emptyset : P(x)\] is TRUE regardless of the value of P(x)

Answer 1

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

Answer 2

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.

Answer 3

But that is like a convention because there is no x to make TRUE the statement.

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

\[x\in \emptyset \Rightarrow P(x)\] totally makes sense to me. What I can't see is how is this equivalent to the universal quantification.

Answer 8

Sorry, can anyone just tell me what \(\rm P(x)\) stands for?

Answer 9

I do know that conditional is not equivalent to \(\rm if\cdots then\) statements.

Answer 10

Oh, I get it.

Answer 11

\[P(x)\] 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 \[x > 0\]

Answer 12

if you give x the value -1 you can say that the proposition is FALSE.

Answer 13

\[\forall x \in \emptyset :\rm P(x)\]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.

Answer 14

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.

Answer 15

If \(x \in \emptyset\), we know that \(P(x)\) is true so the conditional is true.

Answer 16

So the statement is always true, and it is given in your original statement that it is true.

Answer 17

Two statements having the same truth tables are equivalent.

Answer 18

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

\[(x \in \emptyset) \rightarrow P(x)\]

Answer 19

Then can I turn every universal quantification into an implication?

Answer 20

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"

Answer 21

yes you can

Answer 22

generally speaking

\[\forall x \in X, P(x)\]

is equivalent to

\[(x \in X) \rightarrow P(x)\]