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

md2perpe
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GGG
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    What have you tried? Where did you get stuck? Is this a homework problem? – Noah Schweber Sep 08 '19 at 20:16
  • $it; is; an; exercise; from ; an ; exam; from; university; of; Athens. At; 1st; i; had; to; ''translate'' (A\Rightarrow B)\wedge (B\Rightarrow A),using; only; A,B,\wedge, \vee ,\sim .Which ; i ; did;, thinking (A\Rightarrow B)\wedge (B\Rightarrow A) equals; to A\Leftrightarrow B.So,; this ; should ; do: (A\wedge B)\vee (\sim A\wedge \sim B) Next; I; had; to; do; the; same; thing; using; only; A,B,\sim ,\vee ,. I; tried; \sim (A\vee B)\vee (\sim A \vee \sim B),but; i; really; believe; something's; not; right...$ – GGG Sep 08 '19 at 20:39
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    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"). – Noah Schweber Sep 08 '19 at 20:41
  • Okay, got it ! Thanks a lot – GGG Sep 10 '19 at 16:26

2 Answers2

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Yes:

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

J. W. Tanner
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    $A\iff B$ means $(A\land B)\lor (\lnot A \land \lnot B)$; get rid of $\land$ using $C\land D\equiv \lnot(\lnot C\lor \lnot D$) – J. W. Tanner Sep 08 '19 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 $\neg$,$\lor$

AgentS
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