Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Are the 2 equivalent? Please prove your answer

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

2) $∀x(Ax \to Bx)$

share|improve this question
    
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

1 Answer 1

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 $B$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.