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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
Andrew Ryan
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1 Answers1

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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.

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

Graham Kemp
  • 129,094
  • 1
    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}$     – Andrew Ryan Sep 16 '19 at 01:04
  • Yes, indeed. @AndrewRyan – Graham Kemp Sep 16 '19 at 02:33
  • So these are the right steps? what would be the logic then for deducing line 9? – Andrew Ryan Sep 16 '19 at 02:54
  • $\vee$ elimination (1, 3-5, 6-8). – Graham Kemp Sep 16 '19 at 03:28
  • What would be a good resource to learn more about the proof by cases? – Andrew Ryan Sep 16 '19 at 03:55