Follow varying typographic conventions with the same input syntax

Numbers

Delimiters

\documentclass{article}\pagestyle{empty}
\usepackage{mathtools, amssymb, amsfonts}
\begin{document}
$\displaystyle \left( \left( \left( ( ) \sqrt{2} \right) \right) \right)$ \quad
$\displaystyle\delimitershortfall=-1pt \left( \left( \left( ( ) \sqrt{2} \right) \right) \right)$ \qquad
\end{document}

$$\left(\left(\left(()\sqrt{2}\right)\right)\right)\left(\right((()\sqrt{2})\left)\right)$$

\left( \left( \left( ( ) \sqrt{2}  \right) \right) \right) \qquad
\Bigg( \bigg( \Big( ( ) \sqrt{2}  \Big) \bigg) \Bigg)

Fractions

Dynamic macro

\documentclass{article}\pagestyle{empty}
\usepackage{amsmath, lipsum, xcolor}
\let\oldfrac\frac
\renewcommand{\frac}[2]{%
  \mathchoice
    {\oldfrac{#1}{#2}}  % displaystyle
    {^{#1}\!/_{\!#2}} % textstyle
    {\oldfrac{#1}{#2}}  % scriptstyle
    {\oldfrac{#1}{#2}}  % scriptscriptstyle
}
\begin{document}
{\color{gray}\lipsum[1][1-2]}
$\frac{1}{2} + \frac{3}{4} = \frac{5}{4}$
{\color{gray}\lipsum[1][1-2]}
$\tfrac{1}{2} + \tfrac{3}{4} = \tfrac{5}{4}$
{\color{gray}\lipsum[1][1-2]}
\[
\frac{1}{2} + \frac{3}{4} = \frac{5}{4}
\]
\end{document}

Static export

Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac, adipiscing vitae, felis. ${}^1{/}_2+{}^3{/}_4={}^5{/}_4$ Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac, adipiscing vitae, felis. $\frac{1}{2}+\frac{3}{4}=\frac{5}{4}$ Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac, adipiscing vitae, felis.

$$\frac{1}{2}+\frac{3}{4}=\frac{5}{4}$$

\begin{align*}
{^{1}\!/_{\!2}} + {^{3}\!/_{\!4}} = {^{5}\!/_{\!4}} \\
\tfrac{1}{2} + \tfrac{3}{4} = \tfrac{5}{4} \\
\frac{1}{2} + \frac{3}{4} = \frac{5}{4}
\end{align*}

Split Equations to equal part to Multi column

• [p] easy to move equations up and down
• [p] allows tag placement in every line

Dynamic macro

\documentclass{article}
\usepackage{mathtools,amssymb,amsfonts,multicol}
\begin{document}
\begin{multicols}{6}\allowdisplaybreaks\vspace*{-1cm}
\begin{align*}%
    a &= b+c \\
    0 &= t+d \\
    0 &= t+d \\
    0 &= t+d \\
    0 &= t+d \\
    0 &= t+d \\
    0 &= t+d \\
    0 &= t+d \\
    0 &= t+d \\
    0 &= t+d \\
    0 &= t+d
\end{align*}
\end{multicols}
\end{document}

Static export

$$\begin{array}{rlrlrlrlrlrl}a& =b+c& 0& =t+d& 0& =t+d& 0& =t+d& 0& =t+d& 0& =t+d\\ 0& =t+d& 0& =t+d& 0& =t+d& 0& =t+d& 0& =t+d& & \end{array}$$

\begin{align*}
    a &= b+c & 0 &= t+d & 0 &= t+d & 0 &= t+d & 0 &= t+d & 0 &= t+d \\
    0 &= t+d & 0 &= t+d & 0 &= t+d & 0 &= t+d & 0 &= t+d
\end{align*}

Aligned overset

Dynamic macro

\documentclass{article}\pagestyle{empty}
\usepackage{mathtools, amssymb, amsfonts, aligned-overset}
\begin{document}
\begin{align*}\MoveEqLeft{}
    \Big( \exists\, y : \neg \big( P_1(y) \lor P_2(y) \big) \Big) \lor \neg \big( \forall\, z : \neg P_2(z) \lor P_1(z) \big) \to \big( \exists\, x : \neg P_1(x) \big) \\
    \mathrlap{\xLongrightarrow{\hspace{5.5em}}}
    \overset{\text{Proposition 0.5.6}}&{=} 
    \neg\big(\forall\, y : P_1(y) \lor P_2(y)\big) \lor \neg\big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \to \big(\exists\, x : \neg P_1(x)\big) \\
    \overset{\text{Proposition 0.5.6}}&{\equiv} 
    \neg\big(\forall\, y : P_1(y) \lor P_2(y)\big) \lor \neg\big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \to \neg\big(\forall\, x : P_1(x)\big) \\
    \overset{\text{Implikation}}&{\equiv} 
    \neg \Big( \neg\big(\forall\, y : P_1(y) \lor P_2(y)\big) \lor \neg\big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \Big) \lor \neg\big(\forall\, x : P_1(x)\big) \\
    \overset{\text{De Morgan II}}&{\equiv} 
    \Big( \neg\neg\big(\forall\, y : P_1(y) \lor P_2(y)\big) \land \neg\neg\big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \Big) \lor \neg\big(\forall\, x : P_1(x)\big) \\
    \overset{\text{Doppelte Negation}}&{\equiv} 
    \Big( \big(\forall\, y : P_1(y) \lor P_2(y)\big) \land \big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \Big) \lor \neg\big(\forall\, x : P_1(x)\big) \\
    \overset{\text{Kommutativität von $\lor$}}&{\equiv} 
    \neg\big(\forall\, x : P_1(x)\big) \lor \Big( \big(\forall\, y : P_1(y) \lor P_2(y)\big) \land \big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \Big)
\end{align*}
\end{document}

Static export

$$\begin{array}{rl}& (\exists y:\neg (P_1(y)\vee P_2(y)\left)\right)\vee \neg (\forall z:\neg P_2(z)\vee P_1(z))\to (\exists x:\neg P_1(x))\\ & \stackrel{\text{Proposition 0.5.6}}{\equiv }\neg (\forall y:P_1(y)\vee P_2(y))\vee \neg (\forall z:\neg P_2(z)\vee P_1(z))\to (\exists x:\neg P_1(x))\\ & \stackrel{\text{Proposition 0.5.6}}{\equiv }\neg (\forall y:P_1(y)\vee P_2(y))\vee \neg (\forall z:\neg P_2(z)\vee P_1(z))\to \neg (\forall x:P_1(x))\\ & \stackrel{\text{Implikation}}{\equiv }\neg (\neg (\forall y:P_1(y)\vee P_2(y))\vee \neg (\forall z:\neg P_2(z)\vee P_1(z)\left)\right)\vee \neg (\forall x:P_1(x))\\ & \stackrel{\text{De Morgan II}}{\equiv }(\neg \neg (\forall y:P_1(y)\vee P_2(y))\wedge \neg \neg (\forall z:\neg P_2(z)\vee P_1(z)\left)\right)\vee \neg (\forall x:P_1(x))\\ & \stackrel{\text{Doppelte Negation}}{\equiv }\left(\right(\forall y:P_1(y)\vee P_2(y))\wedge (\forall z:\neg P_2(z)\vee P_1(z)\left)\right)\vee \neg (\forall x:P_1(x))\\ & \stackrel{\text{Kommutativita¨t von ∨}}{\equiv }\neg (\forall x:P_1(x))\vee \left(\right(\forall y:P_1(y)\vee P_2(y))\wedge (\forall z:\neg P_2(z)\vee P_1(z)\left)\right)\end{array}$$

\begin{align*} \qquad&\hspace{-2em}
\Big( \exists\, y : \neg \big( P_1(y) \lor P_2(y) \big) \Big) \lor \neg \big( \forall\, z : \neg P_2(z) \lor P_1(z) \big) \to \big( \exists\, x : \neg P_1(x) \big) \\&
\overset{\mathclap{\text{Proposition 0.5.6}}}{\equiv}\hspace{2em} \neg\big(\forall\, y : P_1(y) \lor P_2(y)\big) \lor \neg\big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \to \big(\exists\, x : \neg P_1(x)\big) \\&
\overset{\mathclap{\text{Proposition 0.5.6}}}{\equiv}\hspace{2em} \neg\big(\forall\, y : P_1(y) \lor P_2(y)\big) \lor \neg\big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \to \neg\big(\forall\, x : P_1(x)\big) \\&
\overset{\mathclap{\text{Implikation}}}{\equiv}\hspace{1.3em} \neg \Big( \neg\big(\forall\, y : P_1(y) \lor P_2(y)\big) \lor \neg\big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \Big) \lor \neg\big(\forall\, x : P_1(x)\big) \\&
\overset{\mathclap{\text{De Morgan II}}}{\equiv}\hspace{1.5em} \Big( \neg\neg\big(\forall\, y : P_1(y) \lor P_2(y)\big) \land \neg\neg\big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \Big) \lor \neg\big(\forall\, x : P_1(x)\big) \\&
\overset{\mathclap{\text{Doppelte Negation}}}{\equiv}\hspace{2.3em} \Big( \big(\forall\, y : P_1(y) \lor P_2(y)\big) \land \big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \Big) \lor \neg\big(\forall\, x : P_1(x)\big) \\&
\overset{\mathclap{\text{Kommutativität von $\lor$}}}{\equiv}\hspace{2.9em} \neg\big(\forall\, x : P_1(x)\big) \lor \Big( \big(\forall\, y : P_1(y) \lor P_2(y)\big) \land \big(\forall\, z : \neg P_2(z) \lor  P_1(z)\big) \Big)
\end{align*}

More

$$\begin{array}{rl}[\epsilon {]}_{\equiv A}& =\{\epsilon \}\\ [0{]}_{{\equiv }_A}& =\{x0\mid x\in {\Sigma }^{\ast }\wedge {\left|x\right|}_0\mathrm{mod}2=1\}\\ [01{]}_{{\equiv }_A}& =\{x1\mid x\in {\Sigma }^{\ast }\wedge {\left|x\right|}_0\mathrm{mod}2=1\}\\ [00{]}_{{\equiv }_A}& =\{x0\mid x\in {\Sigma }^{\ast }\wedge {\left|x\right|}_0\mathrm{mod}2=0\}\\ [1{]}_{{\equiv }_A}& =\{x1\mid x\in {\Sigma }^{\ast }\wedge {\left|x\right|}_0\mathrm{mod}2=0\}\\ [11{]}_{{\equiv }_A}& =\{x11y\mid x,y\in {\Sigma }^{\ast }\}\end{array}$$ $$\begin{array}{rl}[\epsilon {]}_{{\equiv }_A}& =\{b^n\mid n\in \mathbb{N}\}\\ [a^k{]}_{{\equiv }_A}& =\{w\in \{a,b{\}}^{\ast }\mid {\left|w\right|}_a=k\}\text{ fu¨r }k\in {\mathbb{N}}^{+}\\ [c{]}_{{\equiv }_A}& =\{xcy\mid x,y\in \{a,b{\}}^{\ast }\wedge {\left|x\right|}_a-{\left|y\right|}_a=0\}\\ [a^{l}c{]}_{{\equiv }_A}& =\{xcy\mid x,y\in \{a,b{\}}^{\ast }\wedge {\left|x\right|}_a-{\left|y\right|}_a=l\}\text{ fu¨r }l\in {\mathbb{N}}^{+}\\ [cc{]}_{{\equiv }_A}& =\{w,xcy\mid w\in \{a,b,c{\}}^{\ast }\wedge {\left|w\right|}_c\ge 2\wedge x,y\in \{a,b{\}}^{\ast }\wedge {\left|y\right|}_a>{\left|x\right|}_a+2\}\end{array}$$ $$\begin{array}{rl}{P}_1:S& \to b\mid abbb\mid acbbb\mid aAbbbbb\mid aACbbbbb\mid aACCbbbbb\\ A& \to AACbb\mid AAbb\\ bC& \to Cb\\ aA& \to aa\\ aC& \to ac\\ cC& \to cc\end{array}$$ $$X=\sum _{1\le i\le j\le n}{X}_{ij}X=\sum _{1\le i\le j\le n}{X}_{ij}$$