each $X_i$. These two base points must coincide, otherwise no nontrivial section in $H^0(X_i,(K_X+\Lambda)|_{X_i})$ can be glued into a section of $H^0(X,K_X+\Lambda)$.

\[decrease1\] Let $(X,\Lambda)$ be a connected Gorenstein stable log surface which has two irreducible components $X_1$, $X_2$. Assume further $\Delta(X_1,(K_X+\Lambda)|_{X_1})=\Delta(X_2,(K_X+\Lambda)|_{X_2})=0$ and $\Delta(X,K_X+\Lambda)=1$. Then $$r_{X_1\to X_1\cap X_2}((K_X+\Lambda)|_{X_1})=r_{X_2\to X_1\cap X_2}((K_X+\Lambda)|_{X_2})=3.$$

By (\[deltaIneq\]), we have $\max\{r_{X_1\to X_1\cap X_2}((K_X+\Lambda)|_{X_1}),r_{X_2\to X_1\cap X_2}((K_X+\Lambda)|_{X_2}\}\le 3$. Moreover $r_{X_i\to X_1\cap X_2}((K_X+\Lambda)|_{X_i}\ge 2$ by Lemma \[restdim\]. Thus there must be one $r_{X_i\to X_1\cap X_2}((K_X+\Lambda)|_{X_i}=3$.

Now we suppose $r_{X_1}(X_1\cap X_2)=2$ and $r_{X_2}(X_1\cap X_2)=3$ for a contradiction. Then $X_1\cap X_2$ is a line on $X_1$ and not a line on $X_2$. Thus nonzero elements of $\mathrm{im}\mathrm{Res}_{X_i|X_1\cap X_2}$ have different degrees. Hence only those sections in $H^0(X_i,(K_X+\Lambda)|_{X_i})$ vanishing on $X_1\cap X_2$ can be glued together. Therefore $$\begin{aligned}
  p_g(X,\Lambda)&=\dim\ker \mathcal{R}_{X_1\to X_1\cap X_2}+\dim \ker \mathcal{R}_{X_2\to X_1\cap X_2}\\
  &\le p_g(X_1,(K_X+\Lambda)|_{X_1})-2+p_g(X_2,(K_X+\Lambda)|_{X_2})-2\\
  &\le (K_X+\Lambda)^2,
  \end{aligned}$$ a contradiction. This completes the proof.

\[diffdelta\] Let $(X,\Lambda)$ be a connected Gorenstein stable log surface. Assume $X=X_1\cup X_2$, where $X_i$ is irreducible. Assume further $\Delta(X,K_X+\Lambda)= \Delta(X_1,(K_X+\Lambda)|_{X_1})=1$ and $\Delta(X_2,(K_X+\Lambda)|_{X_2})=0$.

Then

-   either each $(K_X+\Lambda)|_{X_i}$ is very ample and on each $X_i$ the curve $X_1\cap X_2$ is a line on each $X_i$. $K_X+\Lambda$ is very ample;

-   or $X_1$ is non-normal. $X_1\cap 