Proving a Tautology by Using Logical Equivalences

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Thank you so much.... This was very helpful

joonsoftlips
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If it was q-->p do we say it's the same as p-->q just found out it's going to be ~q ٧ p

halo_jk
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show that the conditional statement (p∨q)∧(¬p∨r)⇒(q∨r) is tautology without using truth tables, Sir can you please solve this question, its urgent

ashmirakhan
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Awesome video. Easy-to-follow and great explanation

epkademy
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Thank you so much for this video. Tomorrow I have an exam and I was supper nervous. And your video just save me😊
Keep up the good work✨
You should get a job in my college🥺

tamayasara
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Many thanks for this help. I kept keeping stuck with this exercise, it is nice to see now how to work my way out of these situations. I appreciate this.

valeriereid
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Rodriguez Helen Williams Richard Taylor Ronald

GaryJacksonMoore-kq
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why do we have professors when people like u

dbuc
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White Sharon Rodriguez Charles Walker Elizabeth

JomilaHak-bu
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Thanks for this video am sitting fory exams on 21 july which is next week 2022 and i was nervous about propositoons proofs and you have saved me.Please continue to make more videos may GOD BLESS YOU🙏🙏

lynamhango
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Thank you sir for using this method. I used modus ponens and Modus tollens to solve these types of problems.

pinku
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Gonzalez Steven Taylor Shirley Lewis Steven

jonbootclifford
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Tnxs for this video it helped me too from #Ethiopia

juicewrldforever
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Lopez Michelle Clark Paul Wilson Frank

MaxPenelope-wj
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where can i find more example problems to practice with

law
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Anderson Brenda Walker Ronald Taylor Paul

HanifaMoni-zc
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Hall Scott Miller Elizabeth Moore Carol

MaxPenelope-wj
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you're a great teacher, thanks Jason

felifrit
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Hi Jason thanks for this explanation. It is very clear. I do know if you still read comments but I want to ask anyone here but I wonder if someone can explain the conditional part. The negation of the left and right. I got stuck here and I found your vid!

finback
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can anyone plz explain the 4th and 5th line, , , , absorption law....?😔😔😔

StayWithMath-pvdx