Proving absorption law for sets

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[livid crag]

converted it to logic but now im not sure how to distribute this

[naive grotto]

make a truth table...

or just know that $(P\lor\qty(P\land Q)) \iff P$

[dry nacelle]

Pterodactyl

[naive grotto]

as $(P\land Q) \implies P$

[dry nacelle]

Pterodactyl

[livid crag]

i know how to do it that way but i need to learn how to prove it using logic

[naive grotto]

truth tables are logic

[naive grotto]

this is it

[livid crag]

i mean logic identities

[naive grotto]

oh

$P\lor\qty(P\land Q) \iff \qty(P\lor P)\land\qty(P\lor Q)\iff P\land \qty(P\lor Q)\iff P$

[dry nacelle]

Pterodactyl

[naive grotto]

$A\cup\qty(A\cap B)=\qty(A\cup A)\cap\qty(A\cup B) = A\cap\qty(A\cup B)=A$

[dry nacelle]

Pterodactyl

[livid crag]

is it not possible to convert it to logic then use logic identities?

so starting from this

[naive grotto]

tell them you made up a new identity

and prove it with truth tables

[livid crag]

bruh

[naive grotto]

via boolean algebra tho

show than x+xy=x

x+xy = x(1+y)=x*1=x

replicate this in logic/set theory

@livid crag

[dusky estuary]

Maybe this would help

[naive grotto]

this isnt logical identities like o wants

[honest snow]

You already gave them the relevant logic identities

All that's left is to translate it to set notation

${x: x\in A \lor x\in (A\land B)} \iff {x:(x\in A \lor x\in A)\land (x\in A \lor x\in B)} \ \iff
{x:x\in A\land (x\in A \lor x\in B} \iff {x:x\in A}$

[dry nacelle]

ShiN

[naive grotto]

this?

[naive grotto]

the op didn't accept this either

[livid crag]

heres an example of what my prof wants

[honest snow]

See my answer

[livid crag]

what property did you use to get from 1st to 2nd step

[honest snow]

Distributivity of or over and

[livid crag]

i see

what about 2nd to 3rd?