# Natural deduction Proof [(p →r) ∨ (q→r)]

Trying to go from [(p →r) ∨ (q→r)] to prove (p∧q)→r.

Wanted to know if I am heading in the right direction with my deductions or where I am getting messed up.

1. (p→r) ∨ (q→r) premise
2. p assumption
3. p→r assumption
4. r →elim 2,3
5. q assumption
6. q→r assumption
7. r →elim 5,6
8. p∧q ∧intro 2,5
9. (p∧q)→r →intro 3-4,6-7,8
• Try to use $p\implies q \Leftrightarrow \neg p \vee q$. I am nota sure about 8. Commented Sep 15, 2019 at 23:48

Wanted to know if I am heading in the right direction with my deductions or where I am getting messed up.

You are not. $$p\wedge q$$ should not be derived from assumptions of $$p$$ and $$q$$, it should be the assumption.

Always keep an eye on the goal.

You wish to prove $$(p\wedge q)\to r$$ from the premise $$(p\to r)\vee (q\to r)$$.

Therefore assume $$p\wedge q$$ aiming to derive $$r$$. To derive that, use a proof by cases.

$$\begin{array}{|l}(p\to r)\vee(q\to r)\\\hline\begin{array}{|l} p\wedge q\\\hline\begin{array}{|l}p\to r\\\hline\vdots\\ r\end{array}\\\begin{array}{|l}q\to r\\\hline\vdots\\r \end{array}\\r\end{array}\\(p\wedge q)\to r\end{array}$$

• So utilizing a proof by cases it would look something more like this? $\begin{array}{|l}(p\to r)\vee(q\to r)\\\hline\begin{array}{|l} p\wedge q \quad assumption\\\hline\begin{array}{|l}p\to r \quad assumption\\\hline p \quad \wedge elim 2\\r\quad \to elim3,4\end{array}\\\begin{array}{|l}q\to r \quad assumption\\\hline q \quad \wedge elim 2\\r \quad \to elim 6,7\end{array}\\r\end{array}\\(p\wedge q)\to r \quad \to intro 2-9\end{array}$ Commented Sep 16, 2019 at 1:04
• Yes, indeed. @AndrewRyan Commented Sep 16, 2019 at 2:33
• So these are the right steps? what would be the logic then for deducing line 9? Commented Sep 16, 2019 at 2:54
• $\vee$ elimination (1, 3-5, 6-8). Commented Sep 16, 2019 at 3:28
• What would be a good resource to learn more about the proof by cases? Commented Sep 16, 2019 at 3:55