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How do I solve the next exercise using inference laws, This Rule is also called, "Constructive Dilema"

\begin{align} p & \rightarrow q \\ -p & \rightarrow -q \\ p & \lor -p \\ \hline \\ q & \lor -q \end{align}

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    $\begingroup$ Which inference laws do you have to work with? $\endgroup$ – Bram28 Nov 28 '16 at 0:19
  • $\begingroup$ @Bram28 I have to use the basic ones, and basic replacement Modus ponens / Modus tollens Biconditional introduction / elimination Conjunction introduction / elimination Disjunction introduction / elimination Disjunctive / Hypothetical syllogism Associativity Commutativity Distributivity Double negation De Morgan's laws Transposition Material implication Exportation Tautology Negation introduction $\endgroup$ – JuanMuñoz Nov 28 '16 at 0:28
  • $\begingroup$ how is your disjunction elimination defined? $\endgroup$ – Bram28 Nov 28 '16 at 0:31
  • $\begingroup$ Do you have conditional introduction, and if so, how is that defined? Also,how is Material implication defined? And finally, what is this 'Tautology' you mention? $\endgroup$ – Bram28 Nov 28 '16 at 0:33
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    $\begingroup$ Anyway, you have to understand that there are many different systems of logic. So, when you say 'use the basic rules of logic' I really don't know what rules I can use, so you have to let me know how all these rules are ddefined in your system. You gave me the definition for Material Implication before, that's good and helpful. Can you also give me the definitions for some of the others, especially disjunctive elimination, conditional introduction (if you hae that one), and 'Tautology'? $\endgroup$ – Bram28 Nov 28 '16 at 0:50
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Here is a proof that just uses the first two premises:

  1. $p \rightarrow q$ Premise

  2. $\neg p \rightarrow \neg q$ Premise

  3. $\neg p \lor q$ Conditional 1

  4. $q \lor \neg p$ Commutative 3

  5. $\neg \neg q \lor \neg p$ Double Negative 4

  6. $\neg q \rightarrow \neg p$ Conditional 5

  7. $\neg q \rightarrow \neg q$ Transitivity 2,6

  8. $\neg \neg q \lor \neg q$ Conditional 7

  9. $q \lor \neg q$ Double Negative 8

And here is a (much simpler!) proof that uses just the third premise:

  1. $p \lor \neg p$ Premise

  2. $\top$ Negation 1

  3. $q \lor \neg q$ Negation 2

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  • $\begingroup$ That means that if you dont have the third premise of my question, you still can obtain the answer?? $\endgroup$ – JuanMuñoz Nov 28 '16 at 1:32
  • $\begingroup$ Correct! In fact, notice that the third premise is a tautology, so that isn't very useful (a tautology has no information content). $\endgroup$ – Bram28 Nov 28 '16 at 1:40
  • $\begingroup$ So our conclusion is too a tautology, and does'nt need premises for being demonstrated???? $\endgroup$ – JuanMuñoz Nov 28 '16 at 1:41
  • $\begingroup$ Yes, typically when the goal is a tautology, you don't need any premises ... But your system is a little unusual in that it cannot infer anything starting with nothing. So, in this case we did need our premises ... Though probably we could have done it using just 1 premise now that I think about it. $\endgroup$ – Bram28 Nov 28 '16 at 1:44
  • $\begingroup$ @JuanMuñoz Just noticed that you can do a much simpler proof ... only using premise 3! See my Answer. $\endgroup$ – Bram28 Nov 28 '16 at 1:54

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