Proof and Problem Solving - Sets Example 05

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The same sets as investigated in Proof and Problem Solving - Sets Example 04 are also examined here. However, instead of establishing set equality using Venn diagrams, we use logical symbols and basic definitions to verify equivalence.
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[@jeremyharris246]

Thanks Adam. It's really helped seeing the laws being used and well explained other than trying to find the patterns straight from a list of laws. It's helpful following you in my notebook and pausing the video trying to guess what comes next. Really appreciate the videos.

[@AdamPanagos]

Glad I could help; thanks for watching!

[@DaiMoscv]

I have been learning different types of proofs but was always hesitated, after this video my mind is clear as the sky.

[@AdamPanagos]

Thanks for the kind words, I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks much, Adam

[@janicepearlsoria1213]

love this explanation,...very helpful...thank you, keep it up...God bless

[@fa6m]

والله يو ار ذا بيسست شرحك ناار انته احسن واحد بالعالم ليش ماتيي تدرسني بجامعتنا لييششش

[@AdamPanagos]

Thanks much, glad I could help!

[@katiebell1138]

Thanks so much for this excellent explanation it helps tremendously. Just wondering if this is all I would need to do in a test or do I need to prove both sides since it is an equality.

[@yuvaliko]

Thank you very much!

[@omarel-ghezawi6466]

At time 4:00 of the video, isn't better to say that the first term is the empty set instead of contradiction?

[@AdamPanagos]

No, those are logical statements so they evaluate to some logical T or F. They aren't sets.

[@maryam-mishal6951]

Thank you so much for this! Really helped :)

[@wangkangcheng]

For the second example, b) U should consider the cases and not "jump" a huge step to show A or ( B and C ) = (A and B) or (A and C)

[@AdamPanagos]

Not sure I follow your comment. This example establishes the equality of two different sets by showing each set is a subset of the other. We do this by picking an arbitrary element of the first set, and logically deduce (using basic definitions) it is in the second set (and then vice versa). Your comment written above is written in terms of logical operators AND/OR, and not set operations (e.g set union or intersection). Also, even if these logical symbols are replaced with similar set operators, the equation you provide is not what is considered in this problem. We establish that A U (B INT C) = (A U B) INT (A U C) in this problem.

[@I3uzzzzzz]

dude ur lit. ty fam

[@bravek9881]

Thank u so much but could you help me in that e.g :
Prove that
((A\B)\(B\C))=(A\B) ???? [HELP]

[@tommcgill8601]

I don't know how to rearrange this nicely in the quickest way but I can prove it and maybe you can expand: x ∈ A\B = x ∈ A and x ∉ B. If x ∈ A then x ∈ A\B, if x ∉ B then x ∉ B\C. so
((x ∈ A\B )\(x ∈ B\C)) = x ∈ ((A\B)\(B\C))

[@rovaniainaraveloson5622]

wouldn't you need to prove the other direction for the last example? Assuming it is a double inclusion proof we would need to show the other direction.

[@AdamPanagos]

No, this isn't a "double inclusion" proof. I started with one expression, and showed that it was equal to the other using logical expression manipulation. Since they're equal we're done. If I'd started with, "Let x be in the first set ....." and showed that it was in the 2nd set, then I'd need to do the "other direction". But, that's not what was done here. Hope that helps.

[@rovaniainaraveloson5622]

@@AdamPanagos copy

[@stevencheng7997]

this exact problem showed up on my test, LOL!

[@AdamPanagos]

Hmm.....very curious.....

[@thunderskull258]

I want for cbse

[@snehalata5358]

u did wrong in 2 example
you proved distributive law by using distributive law....

[@AdamPanagos]

No, I don't think I made a mistake.
In the part (b), the goal was to prove an identity that involved the UNION and INTERSECTION of sets. To accomplish this I used the distributive law of the DISJUNCTION operator (i.e. the OR operator). The distributive law of the disjunction operator was already established previously. The strategy here was to transform the set union/intersection expression into an equivalent notation so we can use core properties of the and/or logical connective to verify the expression. No mistakes made as far as I can tell.