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Example:
1. P ∧ ¬Q : Premise
2. P → Q : Premise
3. P : Conclusion
4. ¬Q : 1, simplification
5. ¬P → ¬Q : 2,4 Implication Equivilence
6. P : 3,4 Modus tollens

This was my original attempt at the question. But I was thinking if its possible to just negate any line could I not do this?

1. P ∧ ¬Q : Premise
2. P → Q: Premise
3. P : Conclusion
4. ¬Q : 1, simplification
5. ¬P: 2,4, Modus tollens
6. P: 5 Negation

Furthermore With the first example if the question asked to prove the argument only using rules of inference would I not be able to then use "Implication Equivalence" as a step?

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  • $\begingroup$ What is "implication equivalence" ? From $P \to Q$ you can derive either $\lnot Q \to \lnot P$ by Contraposition or $\lnot P \lor Q$ by Material Implication. $\endgroup$ – Mauro ALLEGRANZA Jan 8 '18 at 16:22
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    $\begingroup$ Why not simply use the Simplification rule on 1 to derive $P$ ? $\endgroup$ – Mauro ALLEGRANZA Jan 8 '18 at 16:29
  • $\begingroup$ @Mauro ALEGRANZA I thought you had to use all your premise's to derive the conclusion? $\endgroup$ – CalciumTablet Jan 8 '18 at 16:33
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    $\begingroup$ NO, not necessarily. If $\varphi$ follows from a set of premises $\Gamma$, then it fallows also from every "larger" set $\Gamma'$ such that: $\Gamma \subseteq \Gamma '$. $\endgroup$ – Mauro ALLEGRANZA Jan 8 '18 at 16:36
  • $\begingroup$ One nitpick on top of everything else: you probably shouldn't write your desired conclusion as a line in your proof as in line 3 here (until the end when you've actually proved it of course). It is confusing cause normally every line in an a proof is something that we've proven or something that we're assuming for the sake of argument. $\endgroup$ – spaceisdarkgreen Jan 8 '18 at 16:44
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Your premises are inconsistent, and from inconstent premises, you may prove anything at all.

Noteably, if you wish to prove $P$, you just need the first premise and "simplification".

$\qquad\begin{array}{r|l:l} 1 & P\wedge \neg Q & \textsf{Premise 1} \\ \hline 2 & P & 1, \textsf{Conjunction Elimination} \end{array}$

If you wish to prove $\neg P$, you may use Modus Tolens ("deny the consequence") with both premises.

$\qquad\begin{array}{r|l:l} 1 & P\wedge \neg Q & \textsf{Premise 1} \\ 2 & P\to Q & \textsf{Premise 2} \\ \hline 3 & \neg Q & 1, \textsf{Conjunction Elimination} \\ 4 & \neg P & 3,2, \textsf{Conditional Elimination (Modus Tolens)} \end{array}$


And, of course, combining these proofs will prove the inconsistency.

$\qquad\begin{array}{r|l:l} 1 & P\wedge \neg Q & \textsf{Premise 1} \\ 2 & P\to Q & \textsf{Premise 2} \\\hline 3 & \neg Q & 1, \textsf{Conjunction Elimination} \\ 4 & P & 1, \textsf{Conjunction Elimination} \\ 5 & \neg P & 3,2,\textsf{Conditional Elimination (Modus Tolens)} \\ 6 & \bot & 4,5,\textsf{Contradiction Introduction} \end{array}$

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  • $\begingroup$ Sorry; I'm not an English-native. What do you mean with "prove anything at all" ? Prove all or prove nothing ? $\endgroup$ – Mauro ALLEGRANZA Jan 9 '18 at 13:54
  • $\begingroup$ I mean that: when given inconsistent premises you will always be able to make a valid argument to reach any conclusion you wish. An argument is deemed to be valid if the conclusion can not be false when all the premises are simultaneously true. As such, the validity of the argument will be vacuously true when its premises can never be simultaneously true. $\endgroup$ – Graham Kemp Jan 10 '18 at 2:53
  • $\begingroup$ Ok, thanks..... $\endgroup$ – Mauro ALLEGRANZA Jan 10 '18 at 7:11
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It's not generally possible to just negate any line. Rules are inference are supposed to model logical thought patterns; we would not logically proceed from the claim "Asia is east of Europe" to the claim "It is not the case that Asia is east of Europe."

I'm also not sure what your use of "Implication Equivalence" is. It's a term I'm not familiar with. Where did you learn it, and what are the rules of its use? It's also not necessary in the argument -- you don't use line 5 in concluding line 6.

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  • $\begingroup$ Implication Equivalence is the rule that $A\to B~\equiv~\neg A\vee B$.... which CalciumTablet did not use at all. It looks as though CT was actually trying to apply Contraposition but did not get that right either. (That would be $P\to Q ~\equiv~ \neg Q\to \neg P$ .) $\endgroup$ – Graham Kemp Jan 9 '18 at 11:27
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Your Implication Equivalence is problematic. $$P\to Q$$ is not equivalent to $$\lnot P\to \lnot Q$$ It is equivalent to $$\lnot Q\to \lnot P$$.

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  • $\begingroup$ Ohhh, okay I see. How did I not notice that, thank you for the clarification!! $\endgroup$ – CalciumTablet Jan 8 '18 at 16:31
  • $\begingroup$ Welcome. Thanks for your attention. $\endgroup$ – Mohammad Riazi-Kermani Jan 8 '18 at 16:36
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  1. ¬P: 2,4, Modus tollens
  2. P: 5 Negation

Given, e.g., "CalciumTablet is not very good at logic" do you think you can infer that therefore "CalciumTablet is very good at logic"? :-)

Something seems badly amiss here in your understanding of negation, and indeed in your understanding of how Modus Tollens works.

If your current textbook's coverage of deduction is puzzling you, then do take a look at another one! Paul Teller's, for example, is freely available here and is unusually clear.

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