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Can anyone please translate this Predicate Logic statement into English? The question is this:

Let N be the set on Natural numbers. Let S={1,8,4} and T={6,8,11}.

Which of the following statements are correct?

$∃ x. ∀ y. (x ∈ T \land y ∈ T) \land (x ≤ y ∧ x ∈ S)$

I read it as:

There exists an x for all y, x is in the set T and y is in the set T and there exists an x that is less than or equal to y and x is in the set S.

The correct answer was False.

I interpreted this as meaning for every single value of y there is a value of x that is less than or equal to y and every value of x is also in the set S.

Is this correct?

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  • $\begingroup$ I believe it should be as follows: There exists a natural number from both sets such that number is less than or equal to all numbers in the set T. $\endgroup$ – A.Riesen Nov 17 '16 at 0:44
  • $\begingroup$ It might help to use the fact that conjunction of statements is commutative to translate it back to English as well $\endgroup$ – A.Riesen Nov 17 '16 at 0:46
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Yes, it is false, indeed.

Witness that $8$ is the only element in the intersection of $S,T$, and that this is greater than $6$, an element in $T$.   Thus disproving by counterexample.

More so, witness that $1964$ is not in $T$; so not every $y$ is in $T$ even if there exists an $x$ in $S\cap T$.


$∃x.∀y.(x∈T∧y∈T)∧(x≤y∧x∈S)$

"There is some $x$ such that $x$ is in $T$ and $S$ and every $y$ is in $S$ and at least as great as the $x$."

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Don't re-use a quantifier - the $\exists x$ applies to the whole sentence. So the sentence says "there is an $x$ so that for every $y$, $x$ is in $T$ and $y$ is also in $T$ and $x$ is less than or equal to $y$ and $x$ is in $S$." So that means that $x$ has to be in both $S$ and $T$ and must be less than or equal to every $y$, and every $y$ is in $T$. But that clearly can't happen - not every number is in $T$.

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