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I have a question for a deductive reasoning proof i made for the following question: (i needed to prove that these three premises can come to the conclusion of : ~R <--> ~T Im new to this so im not completely sure that this is correct.

Premises 1-3

  1. R--> (S-->T)
  2. S
  3. ~T

End of Premises --------

  1. Sub proof Assumption S
  2. Sub proof T
  3. (end of sub proof) S-->T . -->Intro, 4-5
  4. R . --> Elim 1,6
  5. T -->Elim 2,6
  6. Sub proof Assumption R
  7. Sub proof T
  8. Sub proof ~T
  9. Sub proof ⊥
  10. (end of sub proof) ~R . ~Intro, 9-11
  11. Sub proof ~R
  12. Sub proof ~T
  13. New Sub proof ~T
  14. New sub proof ~R
  15. ~R <--> ~T <--> Intro 13-14, 15-16
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  • $\begingroup$ Your use of $\to$-elim in step 7 is wrong. You cannot derive $R$ from $R \to (S \to T)$ and $S \to T$. $\endgroup$ Dec 6, 2019 at 11:48

1 Answer 1

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We have to derive $\lnot R \to \lnot T$ and vice-versa; then conclude with $\leftrightarrow$-intro.

The first part is straightforward, From 3rd premise : $\lnot T$, using $\to$-intro we get immediately:

4) $\lnot R \to \lnot T$.

For the second part :

5) $R$ --- temporary assumed for a sub-proof

6) $T$ --- from 5) and 2nd premise from 1st one

7) $\bot$ --- contradicition of 6) with 3rd premise

8) $\lnot R$ --- from 5) and 8) by $\lnot$-intro, discharging temporary assumption.

9) $\lnot T \to \lnot R$ --- from 8) by $\to$-intro.

10) $\lnot R \leftrightarrow \lnot T$ --- from 4) and 9) by $\leftrightarrow$-intro.

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  • $\begingroup$ Ok I see now how you got to step 4 but I dont understand much after that in the second part. $\endgroup$ Dec 6, 2019 at 11:53
  • $\begingroup$ Oh I didnt know it was possible to go from premise 1 and 3 directly to the ¬ð‘…→¬ð‘‡. in line 4 $\endgroup$ Dec 6, 2019 at 12:04
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    $\begingroup$ @Michaellaswhich - NO; from premise 3) only. You can use $\to$-intro because you can assume $\lnot R$ and re-iterate $\lnot T$. Then apply $\to$-intro to get $\lnot R \to \lnot T$, discharging $\lnot R$. It can be abbreviated in this way : $\to$-intro allows to derive $A \to B$ from $B$, with $A$ whatever. $\endgroup$ Dec 6, 2019 at 12:09

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