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% (find-angg "LATEX/2016optativa-1.tex")
% (find-angg "LATEX/2016optativa-1.lua")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2016optativa-1.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2016optativa-1.pdf"))
% (defun e () (interactive) (find-LATEX "2016optativa-1.tex"))
% (defun l () (interactive) (find-LATEX "2016optativa-1.lua"))
% (defun u () (interactive) (find-latex-upload-links "2016optativa-1"))
% (defun z () (interactive) (find-zsh "flsfiles-tgz 2016optativa-1.fls 2016optativa-1.tgz")
% (find-xpdfpage "~/LATEX/2016optativa-1.pdf")
% (find-sh0 "cp -v ~/LATEX/2016optativa-1.pdf /tmp/")
% (find-sh0 "cp -v ~/LATEX/2016optativa-1.pdf /tmp/pen/")
% file:///home/edrx/LATEX/2016optativa-1.pdf
% file:///tmp/2016optativa-1.pdf
% file:///tmp/pen/2016optativa-1.pdf
% http://angg.twu.net/LATEX/2016optativa-1.pdf
\documentclass[oneside]{book}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{pict2e}
%\usepackage{tikz}
%
\usepackage{edrx15} % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex % (find-angg "LATEX/edrxaccents.tex")
\input edrxchars.tex % (find-angg "LATEX/edrxchars.tex")
\input edrxheadfoot.tex % (find-dn4ex "edrxheadfoot.tex")
%
\begin{document}
\catcode`\^^J=10
\directlua{dednat6dir = "dednat6/"}
\directlua{dofile(dednat6dir.."dednat6.lua")}
\directlua{texfile(tex.jobname)}
\directlua{verbose()}
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\def\eval#1{\directlua{#1}}
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%\directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end
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% _____ _ _ _
% |_ _|_ _ _ _| |_ ___ | | ___ __ _(_) ___ ___
% | |/ _` | | | | __/ _ \| |/ _ \ / _` | |/ _ \/ __|
% | | (_| | |_| | || (_) | | (_) | (_| | | __/\__ \
% |_|\__,_|\__,_|\__\___/|_|\___/ \__, |_|\___||___/
% |___/
\def\und#1#2{\underbrace{#1}_{#2}}
\def\und#1#2{\underbrace{#1}_{\textstyle #2}}
\def\subf#1{\underbrace{#1}_{}}
\def\p{\phantom{(}}
We can calculate the result of $¬¬P→P$
when $P=0$ (left) and
when $P=1$ (right) with:
\msk
$\und {{\und {¬ {\und {¬ \und P 0} 1}} 0} → {\und P 0}} 1
\qquad
\und {{\und {¬ {\und {¬ \und P 1} 0}} 1} → {\und P 1}} 1
$
\bsk
The {\it subformulas} of $¬¬P→P$ are:
\msk
$\subf{\subf{¬ \subf{¬ \subf P}} → {\subf P}}$
\bsk
If we write the result of each subformula
under its central connective we get:
\msk
$\begin{array}{ccccc}
¬ & ¬ & P & → & P \\ \hline
& & 0 & & 0 \\
& 1 & & & \\
0 & & & & \\
& & & 1 & \\
\end{array}
\qquad
\begin{array}{ccccc}
¬ & ¬ & P & → & P \\ \hline
& & 1 & & 1 \\
& 0 & & & \\
1 & & & & \\
& & & 1 & \\
\end{array}
$
\msk
We can write all results in the same line...
We get something more compact but harder to read,
\msk
$\begin{array}{ccccc}
¬ & ¬ & P & → & P \\ \hline
0 & 1 & 0 & 1 & 0 \\
\end{array}
\qquad
\begin{array}{ccccc}
¬ & ¬ & P & → & P \\ \hline
1 & 0 & 1 & 1 & 1 \\
\end{array}
$
\bsk
We can put each case in a single line.
Here we also add a column at the left with the values of $P$.
\msk
$\begin{array}{ccccccc}
P & & ¬ & ¬ & P & → & P \\ \hline
% (¬\p & (¬\p & \p\p P)) & → & P \\ \hline
0 & & 0 & 1 & 0 & 1 & 0 \\
0 & & 1 & 0 & 1 & 1 & 1 \\
\end{array}
$
\msk
\newpage
% ____ _ _
% / ___|___ _ __ ___ _ __ _ __ ___| |__ ___ _ __ ___(_) ___ _ __
% | | / _ \| '_ ` _ \| '_ \| '__/ _ \ '_ \ / _ \ '_ \/ __| |/ _ \| '_ \
% | |__| (_) | | | | | | |_) | | | __/ | | | __/ | | \__ \ | (_) | | | |
% \____\___/|_| |_| |_| .__/|_| \___|_| |_|\___|_| |_|___/_|\___/|_| |_|
% |_|
Let
$A = \{x:\{-1,...,4\}; x^2\}$ and
$B = \{x:\{-1,...,4\}; x^2≤5; x\}$.
Then $A$ and $B$ can be calculated by:
\msk
$\begin{array}{cc}
x & x^2 \\ \hline
-1 & 1 \\
0 & 0 \\
1 & 1 \\
2 & 4 \\
3 & 9 \\
4 & 16 \\
\end{array}
\qquad
\begin{array}{cccc}
x & x^2 & x^2≤5 & x \\ \hline
-1 & 1 & 1 & -1 \\
0 & 0 & 1 & 0 \\
1 & 1 & 1 & 1 \\
2 & 4 & 1 & 2 \\
3 & 9 & 0 & \\
4 & 16 & 0 & \\
\end{array}
$
\msk
We get:
$A = \{1,0,1,4,9,16\}$,
$B = \{-1,0,1,2\}$.
\bsk
Let
$A = \{x:\{1,...,5\}, y:\{1,...,x\}, x+y≤6; (x,y)\}$ and
$B = \{y:\{1,...,5\}, x:\{y,...,5\}, x+y≤6; (x,y)\}$.
Then $A$ and $B$ can be calculated by:
\msk
$\begin{array}{ccccc}
x & y & x+y & x+y≤6 & (x,y) \\ \hline
1 & 1 & 2 & 1 & (1,1) \\
2 & 1 & 3 & 1 & (2,1) \\
& 2 & 4 & 1 & (2,2) \\
3 & 1 & 4 & 1 & (3,1) \\
& 2 & 5 & 1 & (3,2) \\
& 3 & 6 & 1 & (3,3) \\
4 & 1 & 5 & 1 & (4,1) \\
& 2 & 6 & 1 & (4,2) \\
& 3 & 7 & 0 & \\
& 4 & 8 & 0 & \\
5 & 1 & 6 & 1 & (5,1) \\
& 2 & 7 & 1 & \\
& 3 & 8 & 0 & \\
& 4 & 9 & 0 & \\
& 5 & 10 & 0 & \\
\end{array}
\qquad
\begin{array}{ccccc}
y & x & x+y & x+y≤6 & (x,y) \\ \hline
1 & 1 & 2 & 1 & (1,1) \\
& 2 & 3 & 1 & (2,1) \\
& 3 & 4 & 1 & (3,1) \\
& 4 & 5 & 1 & (4,1) \\
& 5 & 6 & 1 & (5,1) \\
2 & 2 & 4 & 1 & (2,2) \\
& 3 & 5 & 1 & (3,2) \\
& 4 & 6 & 1 & (4,2) \\
& 5 & 7 & 0 & \\
3 & 3 & 6 & 1 & (3,3) \\
& 4 & 7 & 0 & \\
& 5 & 8 & 0 & \\
4 & 4 & 8 & 0 & \\
& 5 & 9 & 0 & \\
5 & 5 & 10 & 0 & \\
\end{array}
$
\msk
We get:
$A = \{ (1,1), (2,1), (2,2), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1)\}$ and
$B = \{ (1,1), (2,1), (3,1), (4,1), (5,1), (2,2), (3,2), (4,2), (3,3)\}$.
\newpage
\end{document}
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lualatex 2016optativa-1
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