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Are the 2 equivalent? Please prove your answer

1) $\forall x(Ax) \to \exists y(By)$

2) $∀x(Ax \to Bx)$

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Thanks to all you beautiful people for your help – user59803 Jan 27 '13 at 2:38
$1$ is equivalent to $\exists x ( Ax \to Bx )$. Is this what you mean? – Andrew Salmon Jan 27 '13 at 2:42
up vote 1 down vote accepted

Let $A(t)$ be the assertion that $t$ is divisible by $4$, and let $B(t)$ be the assertion that $t$ is divisible by $3$. Let our domain be the set $\{1,2,3,4\}$ of integers.

The $\forall x A(x)\to \exists yB(y)$ is true, since $\forall xA(x)$ is false. Actually, it is doubly true, since in fact there is a $y$ such that $B(y)$.

But $\forall x(A(x)\to B(x))$ is false.

The two sentences therefore cannot be logically equivalent.

Remark: There is nothing particularly amusing about integers and divisibility. One can undoubtedly give funnier interpretations of $A$ and $B$.

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What does the 2nd statement translate to in English assuming your integer assertions? – user59803 Jan 27 '13 at 2:26
Every number (in our set) which is divisible by $4$ is divisible by $3$. – André Nicolas Jan 27 '13 at 2:28
On a related note, how is ∀x(A(x) or B(x)) read? – user59803 Jan 27 '13 at 2:51
In my example, it says that every number (in our set) is divisible by $4$ or by $3$ (or both). With the particular choice of numbers I made, this sentence happens to be false. – André Nicolas Jan 27 '13 at 3:09

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