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As I was trying to solve a Tautology, I arrived in the below expression of which I'm confused on how to deal with. My book doesn't explain this logic expression:

$ \lnot p \lor T $

I know the identity for:

$ p \lor T $

Is true, but what about the other one?

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    $\begingroup$ Probably what your book says is that $p\vee T$ is always a tautology for ANY proposition $p$. Notice that if $p$ is a proposition, so is $\neg p$. $\endgroup$ – Darío G Sep 5 '16 at 19:10
  • $\begingroup$ Ok, I see. So if p $\lor T$ is true, $\lnot p \lor T$ is also true. Thank you $\endgroup$ – Nack Sep 5 '16 at 19:15
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$A \lor B$ is true whenever at least one of the disjuncts is true. Since $T$ is by definition always true, so is $\neg p \lor T$.

Convince yourself by drawing a truth table:

| p | ¬p | T | ¬p ∨ T |
|---|----|---|--------|
| 1 | 0  | 1 |    1   |
| 0 | 1  | 1 |    1   |

As you can see, $p \lor T$ is true under all assignments of all variables (here: only $p$).

Additionally, as @Wore said in the comment, $p$ stands as a variable for any proposition, including $\neg p$, i.e. the identity will hold for the proposition's negation as well.

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  • $\begingroup$ Oh ok I see. Being new to logic, I thought the identities shown were valid for p only, and $\lnot$ p would change everything. Thank you. $\endgroup$ – Nack Sep 5 '16 at 19:14
  • $\begingroup$ Possibly it could, you always have to check every assignment to every propositional variable. Just in this case, $\neg p$ doesn't change much since $T$ as the second disjunt still makes the formula true. I've added a truth table for better illustration. $\endgroup$ – lemontree Sep 5 '16 at 19:19
  • $\begingroup$ This helps a lot. Thank you! $\endgroup$ – Nack Sep 5 '16 at 19:21
  • $\begingroup$ What about a statement like $(\lnot p \lor q) \lor T$ This be considered also? $\endgroup$ – Nack Sep 5 '16 at 19:24
  • $\begingroup$ Again, no matter what $(\neg p \lor q)$ evaluates to, $T$ as the second disjunct will always make the formula true. As said by Wore, you can replace the variable $p$ by any proposition and get the same identity. $\endgroup$ – lemontree Sep 5 '16 at 19:27

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