Let $A$ and $B$ be boolean variables. We want to show that

$$\mathrm{\neg }(A\vee B)\leftrightarrow \mathrm{\neg }A\wedge \mathrm{\neg }B.$$

First, assume $\mathrm{\neg }(A\vee B)$ is true. Then, $A\vee B$ is false. If $A$ were true, then we'd have a contradiction, and similarly with $B,$ so both $A$ and $B$ must be false, that is, $\mathrm{\neg }A$ and $\mathrm{\neg }B$ most both be true, and $\mathrm{\neg }(A\vee B)\to \mathrm{\neg }A\wedge \mathrm{\neg }B.$

Now, assume $\mathrm{\neg }A\wedge \mathrm{\neg }B.$ Here, both $\mathrm{\neg }A$ and $\mathrm{\neg }B$ must be true, so both $A$ and $B$ must be false. Therefore, $A\vee B$ is false, so $\mathrm{\neg }A\wedge \mathrm{\neg }B\to \mathrm{\neg }(A\vee B)$ and we have shown $\mathrm{\neg }(A\vee B)\leftrightarrow \mathrm{\neg }A\wedge \mathrm{\neg }B.$

Let $A$ and $B$ be boolean variables. We want to show that

$$\mathrm{\neg }(A\wedge B)\leftrightarrow \mathrm{\neg }A\vee \mathrm{\neg }B.$$

First, assume $\mathrm{\neg }(A\wedge B)$ is true. Then, both $A\wedge B$ must be false, so either $A$ must be false or $B$ must be false, or both must be false. If $A$ is false, then $\mathrm{\neg }A$ is true, and so is $\mathrm{\neg }A\vee \mathrm{\neg }B.$ The same is true if $B$ is false, so $\mathrm{\neg }(A\wedge B)\to \mathrm{\neg }A\vee \mathrm{\neg }B.$

Now, assume $\mathrm{\neg }A\vee \mathrm{\neg }B$ is true. Then, either $\mathrm{\neg }A$ or $\mathrm{\neg }B$ must be true, so either $A$ or $B$ or both must be false. Now, if $A$ is false, then $A\wedge B$ is false. The same holds if $B$ is false, and thus $\mathrm{\neg }(A\wedge B)$ is true. Therefore $\mathrm{\neg }A\vee \mathrm{\neg }B\to \mathrm{\neg }(A\wedge B)$ and we have shown $\mathrm{\neg }(A\wedge B)\leftrightarrow \mathrm{\neg }A\vee \mathrm{\neg }B.$