Proof using Rule of Disjunctive Amplification

[hotvette]

TL;DR

use of negation in disjunctive amplification

Book shows a proof where a conclusion is reached of: $\neg r$. The next step says $\neg r \lor \neg s$ using the rule of disjunctive amplification. The rule of disjunctive amplification as I know it is $p \implies p \lor q$. I don't see how from this you can also say $\neg p \implies \neg p \lor \neg q$. I can see that the truth table is a tautology so I know it's true, I just don't see how to get there.

 

[hotvette]

Is this as simple as letting $p = \neg r$ and $q = \neg s$?

 

[TeethWhitener]

Is this as simple as letting $p = \neg r$ and $q = \neg s$?

Yes.

 

[hotvette]

Really? So this means I can take any of the logic Rules and just replace anything with its negative and vice versa and it is still valid? Example of M. Ponens $[p \land (p \implies q] \implies q$ can be written as $[\neg p \land (\neg p \implies \neg q] \implies \neg q$? The book makes no mention of this. I wonder how we are expected to know...

 

[TeethWhitener]

$p$ is any proposition, including $\neg q$ (kind of. Technically it follows the law of substitution, but functionally it’s the same thing).

 

[TeethWhitener]

Think through it carefully. If you have $\neg p$ and $\neg p \implies \neg q$, why wouldn’t you have $\neg q$?

 

[hotvette]

Sure, that's just M. Ponens. Makes sense, thanks!