Prove this proposition is a tautology: [(p ∨ q) ∧ (p → r) ∧ (q → r)] → r Did I make an error?

Question

Any help is greatly appreciated

[(p ∨ q) ∧ (p → r) ∧ (q → r)] → r ≡

¬[(p ∨ q) ∧ (¬p ∨ r) ∧ (¬q ∨ r)] ∨ r ≡

¬(p ∨ q) ∨ ¬(¬p ∨ r) ∨ ¬(¬q ∨ r) ∨ r ≡ //especially concerned about this step

¬(p ∨ q) ∨ (¬¬p ∧ ¬r) ∨ (¬¬q ∧ ¬r) ∨ r ≡

¬(p ∨ q) ∨ (p ∧ ¬r) ∨ (q ∧ ¬r) ∨ r ≡

¬(p ∨ q) ∨ {(¬r ∧ p) ∨ (¬r ∧ q)} ∨ r ≡

¬(p ∨ q) ∨ {¬r ∧ (p ∨ q)} ∨ r ≡

¬(p ∨ q) ∨ r ∨ {¬r ∧ (p ∨ q)} ≡

¬(p ∨ q) ∨ {(r ∨¬r) ∧ (r ∨(p ∨ q))} ≡

¬(p ∨ q) ∨ {T ∧ (r ∨(p ∨ q))} ≡

¬(p ∨ q) ∨ (r ∨(p ∨ q)) ≡ //also very concerned about last couple steps

¬(p ∨ q) ∨ r ∨ (p ∨ q) ≡

¬(p ∨ q) ∨ (p ∨ q) ∨ r ≡

T ∨ r ≡

T

Answer 1

To address your actual question, the proof you have given is correct. Good job! Could it be better? Sure. The fact that you are "very concerned" about two of the steps indicates to me that you really need to understand why those steps are valid. Is it something about distributivity or associativity combined with DeMorgan? Regardless, do your best to bridge those gaps because your proof is correct--but make sure you understand why. That said, I will present another way of trying to go about it.

---

Start by letting (the reason why will become clear in a second) $$ \Phi\equiv(\neg p\lor r)\land(\neg q\lor r)\tag{1}. $$ And note that $$ \neg(p\lor q)\lor r\equiv(\neg p\land\neg q)\lor r\equiv (\neg p\lor r)\land(\neg q\lor r)\equiv\Phi.\tag{2} $$ Now see if you can follow this reasoning: \begin{align} [(p\lor q)\land(p\to r)\land(q\to r)]\to r &\equiv \neg[(p\lor q)\land(\neg p\lor r)\land(\neg q\lor r)]\lor r\\[0.5em] &\equiv \neg(p\lor q)\lor\neg[(\neg p\lor r)\land(\neg q\lor r)]\lor r\\[0.5em] &\equiv \Phi\lor\neg\Phi\\[0.5em] &\equiv \mathbf{T}. \end{align}

Answer 2

We have $p\lor q$, $p\to r$, and $q\to r$. Now assume $\lnot r$. It follows that $\lnot q \land\lnot p$, that is $\lnot(p\lor q)$, contradicting $p\lor q$.

Answer 3

From your $2^{\text{nd}}$ line: $$(p\lor q)\land(\neg p\lor r)\land(\neg q\lor r)$$ Distribution:

$$(\neg p\lor r)\land(\neg q\lor r)\equiv(\neg p\land\neg q)\lor r\equiv\neg (p\lor q)\lor r$$

We obtain: $$\neg[(p\lor q)\land r]\lor r\equiv\neg(p\lor q)\lor\neg r\lor r\equiv\neg(p\lor q)\lor 1\equiv 1$$ The edited more detailed version with explanation: $$[(p\lor q)\land (p\implies r)\land(q\implies r)]\implies r$$ You've done the following part correctly: $$\equiv\neg[(p\lor q)\land(\neg p\lor r)\land(\neg q\lor r)]\lor r$$ We agree on that. I'll repeat the above: $$(\neg p\lor r)\land(\neg q\lor r)\equiv(\neg p\land\neg q)\lor r\equiv\neg(p\lor q)\lor r$$ Let's return: $$\neg[(p\lor q)\land(\neg(p\lor q)\lor r)]\lor r\equiv$$

$$\neg[((p\lor q)\land\neg(p\lor q))\lor((p\lor q)\land r)]\lor r\equiv$$ $$\neg[0\lor(p\lor q)\land r]\lor r\equiv\neg[(p\lor q)\land r]\lor r$$

Finally: $$\neg(p\lor q)\lor\neg r\lor r\equiv\neg(p\lor q)\lor 1\equiv 1$$