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Anonymous Tue 17 Dec 13:23:02 2019 No.11233070 View Reply Original Report

Quoted By: >>11233683

Anyone knows how this can be resolved while only using boolean algebra (no truth table nor Karnaugh diagrams) ?
I've been trying for hours and still can't find how.

Anonymous

Anonymous Tue 17 Dec 2019 15:42:33 No.11233265 Report

Quoted By:

let b' be the complement of b.
(a+b')*(b+c) = (a+b')*b + (a+b')*c = ab + (bb') + ac + b'c= ab + ac + b'c
but ac<= ab + b'c, since ac*(ab+b'c)'=ac(a'+b')(b+c')=acb'(b+c')=0

Anonymous

Anonymous Tue 17 Dec 2019 18:48:56 No.11233683 Report

Quoted By: >>11233691 >>11234047 >>11234050

>>11233070
truth table?

Anonymous

Anonymous Tue 17 Dec 2019 18:51:12 No.11233691 Report

Quoted By:

>>11233683
OP probably means that since boolean algebras are isomorphic to certain algebras of sets, you can verify a proposition by translating it over to algebras of sets. Hence to prove it, you have to check that for every x, and every possiblity of which sets it belongs it, the proposition is true.

Anonymous

Anonymous Tue 17 Dec 2019 21:22:00 No.11234047 Report

Quoted By:

>>11233683
(a&b)|(~b&c/(a|~b)&(b|c)
https://programming.dojo.net.nz/study/truth-table-generator/index

Anonymous

Anonymous Tue 17 Dec 2019 21:24:18 No.11234050 Report

Quoted By:

>>11233683
(a&b)|(~b&c)/(a|~b)&(b|c)
https://programming.dojo.net.nz/study/truth-table-generator/index

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