Thank you Sir.. Guys may be solving it using digital logic is faster method. dot(.) instead of conjunction and plus(+) instead of disjunction. Use implication as sir said using right associativity. Give it a try guys. and Thank you sir for helping us in many ways.
Sir, please update for Calculus too, if possible.As Engineering Mathematics alone holds major portion it makes us tough to prepare at end and improve it at that moment.
Sir @33:34 in the second question if P is false then P in (P->Q) Should be false (Taking Condition False /\ False = False), If we suppose that (P->Q) to be false then P in (P->Q) has to be true which is not equivalent to the value false of P initially assumed in A, Then it's a Contradiction not Tautology Right??
Hi, if P in P/\ (P->Q) is false then irrespective of any output of (P->Q) the expression P /\ (P->Q) will be false because false /\ is false hence entire LHS is false then entire expression is true
You are a legend. You have the best approach to computational modelling. Cheers
Thank you dave
GATE CS LECTURES BY GATE BOOK u r a genius sir..!!helped me a lot!!...keep it up!
Thank you Sir..
Guys may be solving it using digital logic is faster method. dot(.) instead of conjunction and plus(+) instead of disjunction. Use implication as sir said using right associativity. Give it a try guys. and Thank you sir for helping us in many ways.
Superb explanation.👍👍👍
Sir, please update for Calculus too, if possible.As Engineering Mathematics alone holds major portion it makes us tough to prepare at end and improve it at that moment.
what an amazing explanation Sir.You are great Sir...We want more lectures from you Sir
Salute to you Sir 🤗you saves our lifes,awesome work you do for students,Thanku so much sir
Thank you sir. Your method of teaching is amazing.
Sir how to you got 2^2^n choices.
The way sir says "OKAY" 🤤🥵
your classes are easy to understand sir,thank you
Sir @33:34 in the second question if P is false then P in (P->Q) Should be false (Taking Condition False /\ False = False), If we suppose that (P->Q) to be false then P in (P->Q) has to be true which is not equivalent to the value false of P initially assumed in A, Then it's a Contradiction not Tautology Right??
Hi, if P in P/\ (P->Q) is false then irrespective of any output of (P->Q) the expression P /\ (P->Q) will be false because false /\ is false hence entire LHS is false then entire expression is true
Still not seen the tutor . I am wordless. How to thanks u!!!.
Thanks sir... Great way you taught
Great Work Sir.
thank you sir..:)
Dronacharya Award is waiting for u...
thank you sir .evach lecture i find out very fruitful for me.
Love you sir #NoHomo
Cant we do it with truth table.?
mini verma Cant use in GATE cause it will take more time than gate allow
Sir hindi m bolne se apke views kam nhi honge balki badhenge kuki zyadatar logo ko english nhi aati so plz hindi bol liya kariye
sir can't speak Hindi.
हां भाई, English में दिक्कत होती है
thank you sir .evach lecture i find out very fruitful for me.
thank you sir .evach lecture i find out very fruitful for me.
thank you sir .evach lecture i find out very fruitful for me.
thank you sir .evach lecture i find out very fruitful for me.
thank you sir .evach lecture i find out very fruitful for me.