because 1 \and 0 = 0 same as 1 * 0 = 0 and 1 \or 0 = 1 same as 1 + 0 = 1

Choosing false = 0 and true = 1 is putting the cart before the horse.

It is equally true that 1*0=0 is the same as false|true=true, and 0+1=1 is the same as true&false=false.

But it is also not true that 1+1=1, so it is probably wrong to equate 'or' with '+'. The operation has the wrong properties.

As someone who sometimes dabbles in electronics, 0 = true makes a lot of intuitive sense to me. You have your pin with an open collector, your pull-up resistor, and “true” (as in, it is true that the transistor is conducting) pulls the voltage to ground, which is 0.

As someone who uses a Unix shell, 0 = true makes a lot of intuitive sense to me.

  $ true; echo $?
  0
  $ false; echo $?
  1

You can interpret 0 and 1 as probabilities. 1 + 1 = 1 in this case makes sense because P(A or B) = P(A) + P(B) - P(A and B). You can interpret "A or B" as a set union and "A and B" as a set intersection. Of course it's easy to draw a three-way correspondence between Boolean arithmetic, the events represented by the empty set and the whole space, and sets within some universe because all the objects are so simple, but these correspondences also generalize well to systems with more than two possible values. The ease of generalizing makes me think it's not just a matter of coincidence or convention that we have 0 <=> false.

You've just moved the point where we make the arbitrary choice to here:

> You can interpret "A or B" as a set union and "A and B" as a set intersection.

{True, False, Or, And} and {False, True, And, Or} are two different naming conventions for the exact same structure: the unique boolean algebra on two elements.

A union B is defined as the set of things that are in A or in B; A intersect B is defined as the set of things that are in A and in B. So I don't really see it as an arbitrary choice.

Actually if you think about Boolean logic as mod 2 arithmetics xor is addition, rather than or.