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I need help understanding the following

  • $A \Rightarrow (B \wedge \neg B) \vdash \neg A$
    • Does it mean $(A \Rightarrow (B \wedge \neg B)) \vdash \neg A$. And can I simplify it to $(A \Rightarrow \perp) \vdash \neg A$
  • $(A \vdash B) \vdash A \Rightarrow B$
    • Does it mean $((A \vdash B) \vdash A) \Rightarrow B$

Are my above interpretation right? With added brackets

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2 Answers

In the common standard usage $A \vdash B$ means that there is a proof in the relevant deductive system of $B$ from the premiss $A$. And the deduction theorem says that if $A \vdash B$ then $\vdash A \Rightarrow B$, where the double arrow is the conditional in the deductive system.

You can't write $(A \vdash B) \vdash (A \Rightarrow B)$ since what is on the left of second turnstile is not standardly a wff of the object language but an abbreviatory symbol added to the metalanguage and so a fortiori there is no formal proof in the system from a premiss of the form $A \vdash B$.

You could write $(A \vdash B) \to\ \vdash (A \Rightarrow B)$, where the single arrow is a conditional in the metalanguage, and the double arrow is the object language conditional. But you can't substitute the turnstile for the single arrow or fuse the two.

(If the turnstile belongs to a sequent calculus, parallel remarks still apply.)

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Regarding the first bullet of your question, if you are working with Natural Deduction [es. Chiswell & Hodges, 2007]:

1) $A \rightarrow (B \land \lnot B)$ --- Assumption

2) $B \land \lnot B$ --- from 1, ($\rightarrow$-elim)

3) $B$ --- from 2, ($\land$-elim)

4) $\lnot B$ --- from 2, ($\land$-elim)

5) $\bot$ --- from 3,4 ($\lnot$-elim)

6) $\lnot A$ --- from 5 ($\lnot$-intro)

so (the Assumption 1) is still active) :

$A \rightarrow (B \land \lnot B) \vdash \lnot A$

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