Making something appear below center

**redsoxfan325** · September 23rd 2009, 06:02 PM

$\frac{\displaystyle\frac{{\color{red}P^x}~\display style\frac{\Gamma}{(P\Rightarrow Q)}}{Q}~\displaystyle\frac{\Gamma}{(Q\Rightarrow R)}}{\displaystyle\frac{R}{(P\Rightarrow R)}~x}$

Is there any way to push down that $P^x$ so that it's sitting right on top of the line (while keeping everything else the same)?

Thanks.

**Stefan_K** · October 10th 2009, 06:34 AM

A quick workaround using \llap:

$\frac{\displaystyle\frac{\qquad\displaystyle\frac{ \Gamma}{\llap{$\color{red}P^x$~}(P\Rightarrow Q)}}{Q}~<br /> \displaystyle\frac{\Gamma}{(Q\Rightarrow R)}}{\displaystyle\frac{R}{(P\Rightarrow R)}~x}$

Stefan

**redsoxfan325** · October 10th 2009, 08:45 AM

Originally Posted by Stefan_K

A quick workaround using \llap:

$\frac{\displaystyle\frac{\qquad\displaystyle\frac{ \Gamma}{\llap{$\color{red}P^x$~}(P\Rightarrow Q)}}{Q}~<br /> \displaystyle\frac{\Gamma}{(Q\Rightarrow R)}}{\displaystyle\frac{R}{(P\Rightarrow R)}~x}$

Stefan

One more question: what if I want to adjust something on the right. I tried qquad and rlap, but I got this:

$\frac{\displaystyle\frac{\Gamma}{(Q\Rightarrow P)\rlap{~~$Q$}}}{P}$

**Stefan_K** · October 10th 2009, 08:49 AM

This way for instance:
$\frac{\displaystyle\frac{\Gamma}{(Q\Rightarrow P)\rlap{~~$Q$}}\qquad}{P}$

Stefan

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