I want to prove $P \vee \neg Q \rightarrow R, P \vdash R$, where $\vdash$ means that $R$ is true given that $P \vee \neg Q \rightarrow R$ and $P$ are true (i.e. $R$ is a valid inference from the premises $P \vee \neg Q \rightarrow R$ and $P$)
As an attempt for a formal proof, I came up with the ff:
- $P \vee \neg Q \rightarrow R$ : Given
- $P$ : Given
- $P \vee \neg Q$ : Using the Addition Inference Rule on (2)
- $R$ : Using Modus Ponens on (1,2)
What I am not sure with my proof is the third step, whereby I used the addition rule on $P$ in order to obtain $P \vee \neg Q$. However, I am not sure if this is allowed since strictly speaking, the Addition Inference Rule states that given $P$, we can rewrite it as $P \vee Q$ for any $Q$. However, in this case, I applied to $P$ the $\vee$ operator on a propositional form with negation (i.e. $\neg Q$) ...
Is this is not valid, is there a method for "adding" propositional forms with negations to $P$?
Yes it is perfectly correct. Disjunction Introduction or Addition, states that for any propositions $X,Y$ if you have $\Omega \vdash X$ then you have $\Omega \vdash X\lor Y$. Remember that if $Q$ is a proposition so is $\neg Q$, so the rule applies.