Able to get till A' ∪ (B ∪ C) with set difference and De Morgan's law, but that's different from the right hand side...
Perhaps the statement is false but also not sure how to disprove...
#Prove/disprove subset
rich marlin
autumn oasisBOT
Able to get till A' ∪ (B ∪ C) with set difference and De Morgan's law, but that'...
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sharp sail
Well for any set A and B A\B= A ∩ Comp(B)
sharp sail
Able to get till A' ∪ (B ∪ C) with set difference and De Morgan's law, but that'...
With Comp(B) standing for the complementary of B
Then you can maybe use Morgan’s formula
jagged cairn
$\overline{A\setminus\qty(B\cup C)}=\overline{A\cap\overline{\qty(B\cup C)}}$
hoary pelicanBOT
Romans 12:19
jagged cairn
im not supposed to throw solutions lmao
jagged cairn
**Romans 12:19**
you can use one of the De Morgan's laws of sets from here onwards @rich marlin
remember the complement of an intersection?
rich marlin
Yes, I got A' ∪ (B ∪ C) with set difference and De Morgan's law, but the right hand side is A' ∩ (B ∪ C)...
See the union and intersect after A' is different.
jagged cairn
Yes, I got A' ∪ (B ∪ C) with set difference and De Morgan's law, but the right h...
and that's why it's disproved.
rich marlin
So I can directly say because it's different, it's disproved? Somehow seems too simple for a 5 marks qns...
jagged cairn
give a counterexample too if u like\
rich marlin
Not sure how counter example for set theory looks like, my prof haven't shown any counter examples etc so far... All he ever said was to prove a subset, just show that everything in the first set is in the second set and that's all...
jagged cairn
i mean, it'd suffice to say that the equality doesn't hold true always since intersection and union are different operations.
a counterexample'd work like: we may find anyy 3 sets A,B,C that do not satisfy the equation, thus it does not hold for all the A,B,C it claims to hold for