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Is it possible for $A\Leftrightarrow B$ to be written using only $A,B,\sim,\vee$? If so, how?

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    $\begingroup$ What have you tried? Where did you get stuck? Is this a homework problem? $\endgroup$ Commented Sep 8, 2019 at 20:16
  • $\begingroup$ $it\; is\; an\; exercise\; from \; an \; exam\; from\; university\; of\; Athens. At\; 1st\; i\; had\; to\; ''translate'' (A\Rightarrow B)\wedge (B\Rightarrow A),using\; only\; A,B,\wedge, \vee ,\sim .Which \; i \; did\;, thinking (A\Rightarrow B)\wedge (B\Rightarrow A) equals\; to A\Leftrightarrow B.So,\; this \; should \; do: (A\wedge B)\vee (\sim A\wedge \sim B) Next\; I\; had\; to\; do\; the\; same\; thing\; using\; only\; A,B,\sim ,\vee ,. I\; tried\; \sim (A\vee B)\vee (\sim A \vee \sim B),but\; i\; really\; believe\; something's\; not\; right...$ $\endgroup$
    – GGG
    Commented Sep 8, 2019 at 20:39
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    $\begingroup$ You should add that information (minus the italics) to the question itself. Incidentally, you're on the right track: $(A\wedge B)\vee(\sim A\wedge\sim B)$ is right. Your error was in how you translated the clauses: "$A\wedge B$" is equivalent to "$\sim(\sim A\vee\sim B)$" ("neither $A$ nor $B$ fails"), and similarly "$\sim A\wedge \sim B$" is equivalent to "$\sim (A\vee B)$" ("neither $A$ nor $B$ holds"). $\endgroup$ Commented Sep 8, 2019 at 20:41
  • $\begingroup$ Okay, got it ! Thanks a lot $\endgroup$
    – GGG
    Commented Sep 10, 2019 at 16:26

2 Answers 2

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Yes:

$\lnot (\lnot A\lor \lnot B) \lor \lnot(A\lor B)$

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    $\begingroup$ $A\iff B$ means $(A\land B)\lor (\lnot A \land \lnot B)$; get rid of $\land$ using $C\land D\equiv \lnot(\lnot C\lor \lnot D$) $\endgroup$ Commented Sep 8, 2019 at 20:08
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Using $\neg (A\oplus B)$ is equivalent to $A\iff B$ : $$ \neg(\neg(A \lor \neg B)\lor \neg(\neg A \lor B))$$ In general any logic function can be achieved using just $\neg$,$\lor$

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  • $\begingroup$ thank you very much $\endgroup$
    – GGG
    Commented Sep 8, 2019 at 20:39

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