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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
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  • $\begingroup$ Try to use $p\implies q \Leftrightarrow \neg p \vee q$. I am nota sure about 8. $\endgroup$ Commented Sep 15, 2019 at 23:48

1 Answer 1

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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.

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

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

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