The lattice of subgraphs of a graph form a co-Heyting Algebra.
This is what written in my textbook. However I doubt it, because if I think as follows, it seems to be a Boolean algebra. So what is wrong in my plot? Could give me a counterexample to show the lattice is not a Boolean but a co-Heyting algebra? (Note: It also forms a Heyting algebra, then collectively a bi-Heyting algebra. But I just talk about the co-Heyting side for simplicity.)
Given a graph $G=(V,E)$ s.t. $E\subseteq V\times V$,
let $G_1=(V_1,E_1)$ and $G_2=(V_2,E_2)$ be subgraphs of $G$, i.e. $V_1,V_2\subseteq V$ and $E_1,E_2\subseteq E$.
Also we assume that graph union: $G_1\cup G_2=(V_1\cup V_2,E_1\cup E_2)$.
The co-pseudo-complement $\sim G_1$ is by definition the smallest element such that $G_1\cup \sim G_1 = G$.
I have a hypothesis that $\sim G_1$ can be equivalently obtained by $\sim G_1 = ( (V \setminus V_1)\cup {\bf proj}_1(E\setminus E_1)\cup {\bf proj}_2(E\setminus E_1), E\setminus E_1)$ $ \cdots (\ast)$.
If my hypothesis is correct, it means that we always have the co-pseudo-complement for every subgraph of the given graph $G$, and therefore, the lattice is a Boolean algebra.
If I consider in this way above, I get a wrong conclusion that the lattice of subgraphs of a graph forms a Boolean algebra. I know that a Boolean algebra is a special case of a Heyting algebra, so the statement in the first line of this question is nothing wrong, but I'm afraid that my conclusion is too strong.
So what is wrong in my idea? Is the hypothesis $(\ast)$ wrong?