Rigorous Proof?

[jovial otter]

Is there a rigorous proof for this or is this something that can be shown only through actually using my brain

[barren compass]

There are absolutely rigorous proofs for these

they're also very intuitive

[jovial otter]

???

how would you go about proving this rigorously

[barren compass]

Using the definitions of sets operations

i.e. $A\cup B = {c \vert c \in A \vee c \in B}$

[steep monolith]

lily

[barren compass]

there are many proofs of those propositions you posted online. you can look at them on stackexchange

[jovial otter]

ok i see

so do you think I could say that

according to the definition of a union

it is defined as all the elements x such that x is in set A, x is in set B. Also, x is in set A and B

but because sets A and B are disjoint then by definition there are no elements in set A intersection B

wait but this just seems like I am just spamming definitions

[barren compass]

the important distinction you're missing is that the definition of set union is "the set of all x that are in either A or B (or potentially both)"

while the definition of set intersection is the set of all x that are in both A and B

in the set union it is never guaranteed that x will be in only one of A or B (unless we have other information showing that A and B are disjoint)

[barren compass]

you asked for rigor

i think you'll find that rigorous math is mostly definition and theorem spam

[jovial otter]

gah

I see

ok I see now

thank you for your help

[barren compass]

no prob

i forget if .close works in the help forum posts

[jovial otter]

idk this is my frist time using this help forum thing

[barren compass]

try using ".close" to mark this as solved

[jovial otter]

.close

[iron kayak]

Solved

Post marked as solved by @jovial otter.

Use .unsolved if this was a mistake.