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How can I write:

$A \to B ∧ B \to A $

as a statement with only '$\to$' and '$\neg$' ?

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  • $\begingroup$ What is $\hat{}$? $\endgroup$ – John Griffin Sep 10 '17 at 16:14
  • $\begingroup$ @JohnGriffin presumably logical and $\land$ \land $\endgroup$ – Kenny Lau Sep 10 '17 at 16:14
  • $\begingroup$ @JohnGriffin I edited my answes. Thanks! $\endgroup$ – Raymond Timmermans Sep 10 '17 at 16:16
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We have:

$\lnot (P \to Q) \equiv (P \land \lnot Q)$

and thus:

$\lnot (P \to \lnot Q) \equiv (P \land Q)$.

Thus, we have to rewite the "and" in $(A \to B) \land (B \to A)$ to get:

$\lnot ((A \to B) \to \lnot (B \to A))$

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$$\begin{array}{cl} & [A \implies B] \land [B \implies A] \\ =& [A \implies B] \land \neg \neg[B \implies A] \\ =& \neg([A \implies B] \implies \neg[B \implies A]) \\ \end{array}$$

Alternatively:

$$\begin{array}{cl} & [A \implies B] \land [B \implies A] \\ =& [A \land B] \lor [\neg A \land \neg B] \\ =& \neg[\neg A \lor \neg B] \lor [\neg A \land \neg B] \\ =& [\neg A \lor \neg B] \implies [\neg A \land \neg B] \\ =& [A \implies \neg B] \implies \neg [\neg A \implies B] \\ \end{array}$$

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  • $\begingroup$ @MauroALLEGRANZA right, corrected. $\endgroup$ – Kenny Lau Sep 10 '17 at 16:29

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