So I have to prove this as a tautology. I've been stuck on this forever and am not sure where to go. I experimented and got this far, and looking for some pointers on where to take it next.
(p → q) → ((p v r) → (q v r)) =
(~p v q) → (~(p v r) v (q v r)) : conditional law
~(~p v q) v (~(p v r) v (q v r)) : conditonal law
(p ^ ~q) v (~(p v r) v (q v r)) : De Morgan's law
(p ^ ~q) v ((~p ^ ~r) v (q v r)) : De Morgan's law
this is where I'm stuck. I feel like I should work the distributive law in, but it's not in the correct format for it
Best Answer
The first few steps might seem a bit mysterious as they were figured out with a Karnaugh map, but the trick here is to make every term include either $p$ or $\neg p$: \begin{align*} &(p \land \neg q) \lor (\neg p \land \neg r) \lor q \lor r \\ &\equiv (p \land \neg q) \lor (\neg p \land \neg r) \lor (\top \land q) \lor (\top \land r) \\ &\equiv (p \land \neg q) \lor (\neg p \land \neg r) \lor ((p \lor \neg p) \land q) \lor ((p \lor \neg p) \land r) \\ &\equiv (p \land \neg q) \lor (\neg p \land \neg r) \lor [(p \land q) \lor (\neg p \land q)] \lor [(p \land r) \lor (\neg p \land r)] \\ &\equiv [(p \land q) \lor (p \land \neg q) \lor (p \land r)] \lor [(\neg p \land r) \lor (\neg p \land \neg r) \lor (\neg p \land q)] \\ &\equiv[ p \land (q \lor \neg q \lor r)] \lor [\neg p \land (r \lor \neg r \lor q)] \\ &\equiv[ p \land (\top \lor r)] \lor [\neg p \land (\top \lor q)] \\ &\equiv[ p \land \top] \lor [\neg p \land \top] \\ &\equiv p \lor \neg p \\ &\equiv \top \end{align*}