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I want to prove the law of excluded middle:

$A$

Thus $B ∨ ¬B$

under Copi's logic system.('A' means any premise. I want to derive the law of excluded middle from any premise, not means no premise) Copi system is constructed by 19 rules.

9 rules of inference: modus ponens(MP), modus tollens(MT), hypothetical syllogism(HS), disjunctive syllogism(DS), constructive dilemma(CD), simplification(Simp), conjunction(Conj), addition(Add), absorption(Abs)

10 rules of replacement: De Morgan's rule(DM), Commutativity(Com), Associativity(Assoc), Distrivution(Dist), Double Negation(DN), transposition(Trans), Material implication(Impl), Material equivalence(Equiv), exportation(Exp), Tautology(Taut)

I do not want to use conditional proof and indirect proof(reductio ad absurdum). Can I prove the above statement using only 19 rules?

Bram28
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Idiot
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    Have you tried it yourself? Where have you gotten, Idiot? – David P Oct 31 '19 at 01:20
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    @DavidPeterson I was taken aback for a moment, till I read OP's username – Rushabh Mehta Oct 31 '19 at 01:23
  • I saw this system in "Introduction to logic"(Irving M.Copi, Carl Cohen), and this system is complete. So there exist a proof fo the above. I tried, but I cannot find. – Idiot Oct 31 '19 at 01:28
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    @Idiot Without CP or IP, it's hard to believe the system is complete, because all $19$ rules infer something from something, so you need to start with something to prove any tautology. I understand that this is the reason you throw in the $A$ as a premise ... but my point is that if the system would be complete, it should be able to prove any tautology from nothing. Unless it uses some notion of completeness I am not familiar with. – Bram28 Oct 31 '19 at 02:44
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    @MauroAllegranza The Copi system is not a Fitch system, so the proof is quite different – Bram28 Oct 31 '19 at 13:49

1 Answers1

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\begin{array}{lll} 1. & A & Premise\\ 2. & A \lor \neg B& Add \ 1\\ 3. & \neg B \lor A& Comm \ 2\\ 4. & B \to A & Impl \ 3\\ 5. & B \to (B \land A) & Abs \ 4\\ 6. & \neg B \lor (B \land A) & Impl \ 5\\ 7. & (\neg B \lor B) \land (\neg B \lor A)& Dist \ 6\\ 8. & \neg B \lor B & Simp \ 7\\ 9. & B \lor \neg B & Comm \ 8\\ \end{array}

Bram28
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