# propositional logic syntax using different symbols

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

• What have you tried? Where did you get stuck? Is this a homework problem? Sep 8, 2019 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 8, 2019 at 20:39
• 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"). Sep 8, 2019 at 20:41
• Okay, got it ! Thanks a lot
– GGG
Sep 10, 2019 at 16:26

## 2 Answers

Yes:

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

• $A\iff B$ means $(A\land B)\lor (\lnot A \land \lnot B)$; get rid of $\land$ using $C\land D\equiv \lnot(\lnot C\lor \lnot D$) Sep 8, 2019 at 20:08

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 $$\neg$$,$$\lor$$

• thank you very much
– GGG
Sep 8, 2019 at 20:39