Visuelle Gleichungslinks in Amsmath-Align-Umgebungen

Question

Ich möchte Gleichungen imalign

Da ein Bild mehr sagt als Worte, hier ein Beispiel, das ich in MS Paint erstellt habe, um zu zeigen, was ich meine:

Wie könnte man das erreichen? Ein MWE zu Testzwecken:

\documentclass{article}
\usepackage{amsmath}
\allowdisplaybreaks

\begin{document}
    First, let us solve the following recursion formula:
        $$ F_{n + 1} = \alpha F_{n} + \beta$$
        \begin{align*}
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t}
        \end{align*}
\end{document}

Answer 1

Eine Lösung gemäß meinem Kommentar:

\documentclass{article}
\usepackage{amsmath}
\allowdisplaybreaks
\usepackage{tikz}
\def\tikzmark#1{\begin{tikzpicture}[remember picture]\coordinate(#1);\end{tikzpicture}}
\begin{document}
 $$ F_{n + 1} = \alpha F_{n} + \beta$$
        \begin{align*}
            &\tikzmark{A}\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\tikzmark{C}\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
          &\tikzmark{D}\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
      &\tikzmark{B}\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t} \\
            &\quad \\
            &\quad F_{n + 1} = \alpha F_{n} + \beta \\
            &\equiv \sum_{n = 0}^{\infty} F_{n + 1} t^{n} = \alpha \sum_{n = 0}^{\infty} F_{n} t^{n} + \beta t^{n} \\
            &\equiv t^{-1} \sum_{n = 0}^{\infty} F_{n + 1} t^{n + 1} = \alpha \sum_{n = 0}^{\infty} F_{n}t^n + \beta t^n \\
            &\equiv \phi(t) - F_{0} = \alpha t\phi(t) + \frac{\alpha t}{1 - \beta t} \\
            &\equiv \phi(t) (1 - \alpha t) =  \frac{\alpha t}{1 - \beta t} + F_0\frac{1 - \beta t}{1 - \beta t}
        \end{align*}
\begin{tikzpicture}[remember picture,overlay]
  \draw[-,red] (A)--([xshift=-0.6cm]A)|-(B);
   \draw[-,blue] (C)--([xshift=-0.4cm]C)|-(D);
\end{tikzpicture}
\end{document}

Ausgabe:

Beachten Sie, dass die Linien in der Mitte jeder Reihe (jeder mathematischen Zeile) beginnen und möglicherweise angepasst werden müssen, um mit \equivdem Symbol zentriert zu sein.

Vielleicht kann ich das später automatisieren, wenn Sie interessiert sind. (Eine yshift=2mmOption vor dem Buchstaben des Tikzmarks im Zeichenbefehl kann das manuell beheben)