Shall we call this relation an "isomorphism"?

Question

This is a short question:

Assume the values: $\mathbf{True}$, $\mathbf{False}$

and the logic symbols: $\land,\lor$

Is $\mathbf{True}$, $\mathbf{False}$ under $\land,\lor \,$ isomorphic to $\boldsymbol{0},\boldsymbol{1}$ under $\cdot, +$?

Can you explain why or why not?

Answer 1

In addition to drhab's comment, even if we use addition mod $2$, there is no isomorphism between the two systems. $x+x=0$ for all $x=0,1$ in modulo $2$, so $x+x$ is a constant function. But $x\wedge x=x$ and $x\vee x=x$ for $x=\mathbf{True},\mathbf{False}$, so both $x\wedge x$ and $x\vee x$ are not constant functions.

What you can do is show that $\{\mathbf{True},\mathbf{False}\}$ along with $\wedge$ is isomorphic to $\{0,1\}$ with $\cdot$. I take the map $f:\{\mathbf{True},\mathbf{False}\}\to\{0,1\}$ with $f(\mathbf{True})=1$ and $f(\mathbf{False})=0$. So, $$f(x\wedge y)=f(x)\cdot f(y).$$ Similarly, you can show that $\{\mathbf{True},\mathbf{False}\}$ along with $\vee$ is isomorphic to $\{0,1\}$ with $\cdot$. I take the map $g:\{\mathbf{True},\mathbf{False}\}\to\{0,1\}$ with $f(\mathbf{True})=0$ and $f(\mathbf{False})=1$. So, $$g(x\vee y)=g(x)\cdot g(y).$$