Reliably remove baseline skip between text and math

Question

I'm using ntheorem, with a proof environment that roughly works like

\newenvironment{proof}{\par\textbf{Proof} }{}

My proofs are just blocks of align, so I want to get rid of the long space between “Proof” and the first line of the equation. This could be done like

\begin{flalign}
&\text{\textbf{Proof} to show: }&  x &= y &\\
&& more & math
\end{flalign}

But the “Proof” text comes from the environment, I can't move it into the flalign. So I tried to just set the first math baseline to the last text baseline which would allow writing:

\documentclass{minimal}
\usepackage{amsmath}
\setlength{\textwidth}{7cm}

\begin{document}
\newlength{\antiskip}
\setlength{\antiskip}{\abovedisplayskip}
\addtolength{\antiskip}{\baselineskip}

% Ok
Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac,
\begin{align}\noalign{\vskip-\antiskip}
    \qquad\text{baseline } x^2
\end{align}

% Breaks
Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac,
\begin{align}\noalign{\vskip-\antiskip}
    \qquad\text{baseline } \sum_{i=0}^n
\end{align}

% Desired outcome (yes, really)
Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac,
\begin{align}\noalign{\vskip-\antiskip\vskip-1.5ex}
    \qquad\text{baseline } \sum_{i=0}^n
\end{align}

\end{document}

TL;DR: I want it to look like (3)

Answer 1

Are you after something like this?

Code

\documentclass{article}

\usepackage{amsmath}
\usepackage[standard]{ntheorem}
\theoremstyle{nonumberplain}
\theorembodyfont{\upshape}
\renewtheorem{proof}{Proof}

\begin{document}
\setlength{\abovedisplayskip}{0pt}
\begin{proof}%
Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac, ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac.

With \texttt{flalign}
\begin{flalign}
a_{11}& =b_{11}&
a_{12}& =b_{12}\\
a_{21}& =b_{21}&
a_{22}& =b_{22}+c_{22}
\end{flalign}

With \texttt{equation}
\begin{equation}%\noalign{\vskip-\antiskip}
    %\qquad\text{baseline } 
    \sum_{i=1}^n (x+y)^n = x^n+\binom{n}{1}x^{n-1}y+\binom{n}{2}x^{n-2}y^2+\cdots+\binom{n}{n-1}xy^{n-1}+y^n
\end{equation}

With \texttt{align}
\begin{align}
f(c_1)\cdot\Delta x&=c_1\cdot\Delta x=0\cdot\frac{8}{n}\\
f(c_2)\cdot\Delta x&=c_2\cdot\Delta x=\frac{8}{n}\cdot\frac{8}{n}\\
f(c_3)\cdot\Delta x&=c_3\cdot\Delta x=2\cdot\frac{8}{n}\cdot\frac{8}{n}\\
f(c_4)\cdot\Delta x&=c_4\cdot\Delta x=3\cdot\frac{8}{n}\cdot\frac{8}{n}\\
\vdots\\
f(c_k)\cdot\Delta x&=c_k\cdot\Delta x=(k-1)\cdot\frac{8}{n}\cdot\frac{8}{n}=(k-1)\cdot\frac{64}{n^2}\\
\vdots\\
f(c_n)\cdot\Delta x&=c_n\cdot\Delta x=(n-1)\cdot\frac{8}{n}\cdot\frac{8}{n}=(n-1)\cdot\frac{64}{n^2}
\end{align}
\end{proof}
\end{document}

Output