my logic professor just assigned our class this question that's got us all stumped. We'd appreciate your help. "We have found 2 translations for 'all logicians other than Aristotle are evil' namely $\forall x((Lx\land \lnot a=x)\to Ex)$ and $\lnot\exists x(Lx\land\lnot E x\land\lnot x=a)$. Use formal proof to determine whether these two translations are equivalent or not. That is, show the inference from one to the other is valid/invalid and vice versa. Give counterexample if invalid. How would we go about doing this proof? Thanks, we appreciate your help!

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    $\begingroup$ What proof system are you using? (Rules of inference, axioms, display format, etc.) $\endgroup$ – Graham Kemp Jun 27 '18 at 15:23
  • $\begingroup$ Are you given a set of equivalence principles, like De Morgan? $\endgroup$ – Bram28 Jun 27 '18 at 15:28
  • $\begingroup$ 1st step) use the equivalence of $\forall$ with $\lnot \exists \lnot$. $\endgroup$ – Mauro ALLEGRANZA Jun 27 '18 at 15:33
  • $\begingroup$ 2nd step) use the propositional equivalence of $\lnot (p \to q)$ with $(p \land \lnot q)$. $\endgroup$ – Mauro ALLEGRANZA Jun 27 '18 at 15:34
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    $\begingroup$ @jzaza: You need to show what the rules in your inference system are (not just recite the names your book is using for them). $\endgroup$ – Henning Makholm Jun 28 '18 at 1:05

the two translations are equivalent: $$1)~ (∀x)((Lx∧¬a=x)→Ex)~~~~~$$ $$2)~ ¬¬(∀x)((Lx∧¬a=x)→Ex)$$ $$3)~ ¬(∃x)¬((Lx∧¬a=x)→Ex)$$ $$4)~ ¬(∃x)¬(¬(Lx∧¬a=x)∨Ex)$$ $$5)~ ¬(∃x)((Lx∧¬a=x)∧¬Ex)~~$$


Hint [ref.to : Harry Gensler, Introduction to Logic (3rd ed., 2017)].

For the direction : $\lnot \exists x (\ldots) \to \forall x (\ldots)$.

1) $¬∃x(Lx∧¬Ex∧¬x=a)$ --- premise

2) $∀x ¬(Lx∧¬Ex∧¬x=a)$ --- from 1) by (RS) [page 246 : rules for exchanging quantifiers]

3) $¬∀x((Lx∧¬a=x)→Ex)$ --- [a] assumed the negation of the conclusion

4) $∃x ¬((Lx∧¬a=x)→Ex)$ --- from 3) by (RS)

5) $¬((Lb∧¬a=b)→Eb)$ --- from 4) by (DE) : Drop existential, with $b$ new

6) $(Lb∧¬a=b)$ and $¬Eb$ --- from 5) by (NIF) [page 196]

7) $¬((Lb ∧ ¬a=b) \land ¬Eb)$ --- assumed [b]

8) $(Lb ∧ ¬a=b)$ and $¬¬Eb$ --- from 7) by (NOT-BOTH)

Now we have a contradiction in 6) and 8); thus, by RAA, we can derive :

9) $((Lb ∧ ¬a=b) \land ¬Eb)$ --- closing the innere sub-proof from assumption [b]

10) $¬(Lb ∧ ¬Eb ∧ ¬b=a)$--- from 2) by DU :Drop universal [page 248], re-using $b$

Now we have a new contradiction in 9) and 10); thus, using RAA again, we can derive :

11) $∀x((Lx ∧ ¬a=x) → Ex)$--- closing the outer sub-proof from assumption [a].

Similar for the direction : $\forall x (\ldots) \to \lnot \exists x (\ldots)$.


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