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Indicate if the following reasoning is valid or not: $$\begin{matrix}\forall x:&i(x)&\wedge&\neg v(x)\\\exists x:&v(x)&\wedge& b(x)\\\hline\exists x:&b(x)&\wedge&\neg i(x)\end{matrix}$$


I don't know if the first part my demonstration is OK: I used the universal elimination and the existencial eliminiation to get $$\begin{matrix}i(a)&\wedge&\neg v(a)\\v(a)&\wedge& b(a)\\\hline b(a)&\wedge&\neg i(a)\end{matrix}$$

Is that OK?

Then I did:

$$[i(a)\wedge\neg v(a)\wedge v(a)\wedge b(a)]\quad\Rightarrow\quad[b(a)\wedge\neg i(a)]\\\left[i(a)\wedge F\wedge b(a)\right]\quad\Rightarrow\quad[b(a)\wedge\neg i(a)]\\F\quad\Rightarrow\quad[b(a)\wedge\neg i(a)]\\\neg(F)\quad\vee\quad[b(a)\wedge\neg i(a)]\\V\quad\vee\quad[b(a)\wedge\neg i(a)]\\V\\\therefore\quad\text{The reasoning is VALID}.$$

Is that OK?

Thank you!

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  • $\begingroup$ To be clear, you need to use existential elimination first, then instantiate the universal quantifier with the produced eigenvariable. $\endgroup$ Jun 5 '18 at 1:23
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You have demonstrated: first eliminate the existential, then eliminate the universal to the same witness, so deriving a contradiction ($v(a)\wedge\neg v(a)$), so that the conclusion may follow by the principle of explosion (ex falso sequitur quodlibet).

Therefore the argument is vacuous, but quite valid

$\def\fitch#1#2{~~~\begin{array}{|l} #1\\\hline #2\end{array}}$ $$\fitch{~1.~\forall x~(i(x)\wedge\neg v(x))\\~2.~\exists x~(v(x)\wedge b(x))}{\fitch{~3.~[a]~v(a)\wedge b(a)\qquad~:2,\exists\mathsf e}{~4.~v(a)\qquad\qquad\qquad~~:3,\wedge\mathsf e_1\\~5.~i(a)\wedge \neg v(a)\qquad\quad:1,\forall\mathsf e\\~6.~\neg v(a)\qquad\qquad\quad~~~:5,\wedge\mathsf e_2\\~7.~\bot\qquad\qquad\qquad\quad~~:4,6,\neg\mathsf e}\\~8.~\exists x~\bot\qquad\qquad\qquad\quad~:3,7,\exists\mathsf i\\~9.~\bot\qquad\qquad\qquad\qquad~~~:8,\exists\mathsf e\\10.~\exists x~(b(x)\wedge\neg i(x))\qquad:9,\textsf{EFQ}}$$

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