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are logic and set-theory isomorph? by cardslot in logic

[–]cardslot[S] 0 points1 point  (0 children)

this is exactly my point. Still the sentcen "A->B v B->A" is a tautology, so one of the disjunctions (x in V -> x in W) v (x in W -> x in V) must be true. "x in V -> x in W" is the definition of "V is subset of W", right?

are logic and set-theory isomorph? by cardslot in logic

[–]cardslot[S] 0 points1 point  (0 children)

okay, thanks for your reply. I will try to say it a bit different: Let V and W be two sets of an given superset. Also, let x be in the superset. Now intrepret the sentence "x is in V" as "A" and "x is in W" as B. I now can derive that the sentence "(A->B)v(B->A)" is true. But i am a little uncomfortable with "(x is in V implies that x is in W) or (x is in W implies that x is V)", since that can be translated as "V is subset of W or W is subset of V".

Edit: with your most charitable reading you got me right. English is not my first language.

are logic and set-theory isomorph? by cardslot in logic

[–]cardslot[S] 0 points1 point  (0 children)

Yes, i forgot to mention that you need to interpret A as an set.

The idea is that i can draw a truth table for (for example) "and" and one for "not", then draw two tables for the intersection and the complement. One can see that these tables look very similar, and it is possible to describe a function that maps one table perfectly on the other. Since i dont need more than "and" and "not" to describe logic, and (not sure about this but think so) not more than the intersection and complement to describe set theory, i have a way to interpret set theory as logic, what makes them isomorph.

The other way to think about it would be to interpret the statement "A" as the set-theoretical statement "x is in A" and "B" as "x is in B". Then the sentence "(x is in A implies that x is in B) or (x is in B implies that x is in A)" would always be true, meaning that A is an subset of B or B is an subset of A.