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I know how to show a CNF formula is a tautology using resolution, I've come across this DNF now and have to show its a tautology, is it the same technique or do I need to change it to a CNF? if so how?

             F := (A ∧ B) ∨ (A ∧ ¬B) ∨ (¬A ∧ B) ∨ (¬A ∧ ¬B)
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  • $\begingroup$ Group the first two and second two terms together. $\endgroup$ – Cameron Williams Jan 19 at 16:51
  • $\begingroup$ You can just use a truth table, and show that $F$ is always True. I can give answer below if needed $\endgroup$ – NazimJ Jan 19 at 16:58
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You can consider a truth table for the statement:

$$\begin{bmatrix} & | & A & \neg A \\ \hline B & | & T \lor F \lor F \lor F = T & F \lor F \lor T \lor F = T \\ \neg B & | & F \lor T \lor F \lor F = T & F \lor F \lor F \lor T = T \end{bmatrix}$$

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  • $\begingroup$ Thanks that makes so much sense $\endgroup$ – bodwm15 Jan 19 at 17:23

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