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Prove that not every boolean function is equal to a boolean function constructed by only using $∧$ and $∨$.

If p,q = (0,1)

(p$∧$q)$∨$q = (0$∧$1)$∨$1 = 1

(p$∧$q)$∨$~q = (0$∧$1)$∨$~1 = 0

Therefore (p$∧$q)$∨$q $≠$ (p$∧$q)$∨$~q.

Is my proof good enough? I realize that this question has been asked more than once here but I didn't quite understand the proofs that they gave. Any feedback will be greatly appreciated thanks!

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  • $\begingroup$ No, it is not. You really have not proved anything related to the question. $\endgroup$ – Andrés E. Caicedo Apr 9 '14 at 6:13
  • $\begingroup$ @AndresCaicedo thanks. Which basically means I don't understand the question. If you're willing can you please explain to me what it is asking me to do? $\endgroup$ – JayS Apr 9 '14 at 6:35
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    $\begingroup$ You need to give an example of a Boolean function $f$ with the property that $f\ne g$ for all Boolean functions $g$ constructed using only $\land$ and $\lor$. $\endgroup$ – Andrés E. Caicedo Apr 9 '14 at 6:41
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    $\begingroup$ You could represent the function f such that f(x)=x using $\land$ and $\lor$ by using [x$\land$(x$\lor$y)]. There's a Boolean function "1" such that 1 (x, y)=1 for all x, y, which can represent every tautology. Can you make any tautologies using just $\land$ and $\lor$ and variables? $\endgroup$ – Doug Spoonwood Apr 9 '14 at 14:42
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I don't think it's good enough because you are just showing that two particular boolean functions are not always equal. Hint: use a single variable, two is unnecessarily complicated.

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