$\wedge,\cap,\times$ and $\vee,\cup,+$ are always interchangeable? 
Update : Should have left the Arithmetic out of this question, the new
  modified question is posted here : $\wedge,\cap$ and $\vee,\cup$
  between Logic and Set Theory always interchangeable?

Although  have not seen this anywhere stated, but so often mentally swapping the frameworks of Logic, Set Theory and Arithmetic following operation $\wedge,\cap,\times$ 
and $\vee,\cup,+$ are seem to be always interchangeable (Isomorphic?).
My question is are $\wedge,\cap,\times$ the same thing and just different symbols are being used depending on the framework? same question regarding $\vee,\cup,+$
PS: In arithmetic setting we get the case of $[0, \text{any number other than 0}] \equiv [false , true] $
For example De Morgan's laws for Sets and Logic just becomes the distributive and associate laws.
What about infinite cases? does this type of intuition break down between Arithmetic, Logic and Set theory?
 A: As requested; (I think the question is a bit confused and fuzzy as it stands, but...)
In set theory, $\cap$ and $\cup$ distribute over each other:
$$a\cap(b\cup c) = (a\cap b)\cup (a\cap c)\qquad\text{and}\qquad a\cup(b\cap c) = (a\cup b)\cap (a\cup c).$$
However, in arithmetic (by which I understand the Peano arithmetic or its extensions to $\mathbb{Z}$, $\mathbb{Q}$, etc), while $\times$ distributes over $+$, $+$ does not distribute over $\times$. We always have $a\times(b+c) = (a\times b)+(a\times c)$, but almost never have $a+(b\times c) = (a+b)\times(a+c)$. 
So $\cap$ and $\cup$ in set theory have properties, relative to one another, that are different from the properties of $\times$ and $+$ in arithmetic.
A: "+" isn't like "$\lor$".  (x+x)=(2*x).  On the other hand, in classical logic, (x $\lor$ x)=x.  Specifically, (1+1)=2, while (1 $\lor$ 1)=1.
It comes as more useful to compare "∧,∩, min" and "∨,∪, max".  Logical and set-theoretic operations behave more like the minimum of two numbers, and the maximum of two numbers than anything else (and max and min are dual operations if we have (1-x) as the complement of x). 
