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-1 votes
2 answers
153 views

How can I tell if this boolean expression is a tautology without proving it?

((X'+ Y)·(Z'+ Y'))' + (Z'+ X') So I proved this to be a tautology. And a follow up question I received was how can I know it's a tautology simply by looking the original boolean expression. I can't ...
uwuwuuwuwuw's user avatar
-1 votes
1 answer
217 views

Simplifying a logical equivalence [closed]

http://www2.ift.ulaval.ca/~dadub100/cours/H09/22257/ntsLogique.pdf If you look in Annex B, I am allowed to use all laws from chapter 3 and below. https://i.sstatic.net/KrCLK.jpg This is the problem ...
Samuel Parent's user avatar
-2 votes
2 answers
45 views

logic - how to convert this formula

I have this formula: $$(X \wedge (Y \rightarrow Z)) \vee \neg(\neg X \rightarrow (Y \rightarrow Z))$$ Is it possible to convert it to this: $$X ↔ (Y → Z)$$ the truth table show that they are ...
Courtney Mill's user avatar
-2 votes
2 answers
62 views

How to form a CNF of following formula [closed]

We got an exercise to make a CNF out of the following formula: $$G = ((A \vee \neg B \vee C) \wedge (C \vee D)) \vee ((A \vee \neg C) \wedge (B \wedge D))$$ I've tried to make an equivalent equation ...
Maxim Glazunov's user avatar
-2 votes
1 answer
361 views

By using laws of algebra of statements show that p or (p and q) is equivalent to p . [closed]

I verified the above result using truth tables and got that the LHS is equivalent to the RHS, but I am not able to prove the result using the laws of the algebra of statements. Please guide me.
Sameer Nilkhan's user avatar
-3 votes
1 answer
77 views

Infinity in logic implicit (compact) notation.

This question is about notation. Is this true? $$\bigvee_{i=1}^{\infty}~p_{i} \Longleftrightarrow \bigvee_{i=1}^{}~p_{i}$$ I mean, will it represent the same if I write $\infty$ or left it blank on ...
vasili111's user avatar
  • 368
-4 votes
1 answer
177 views

Boolean algebra expression simplification in conjuction normal form

Is there a way to simplify this expression: (!a || !c || b) && (!a || b) && (a) && (!a || !b || c) && (!b || !d || a) && (d || !c || !b) && (!d || e) ...
adk's user avatar
  • 15

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