Terminterpretation/((0+1)+x)(1+(y+1))/Verschiedene Strukturen/2/Aufgabe/Lösung

Es ist
${}{\left({\left(0+1\right)}+5\right)}\cdot {\left(1+{\left(3+1\right)}\right)}=6\cdot 5=30\,.$
Es ist
${}{\begin{aligned}{\left({\begin{pmatrix}0&0\\0&0\end{pmatrix}}+{\left({\begin{pmatrix}1&0\\0&1\end{pmatrix}}+{\begin{pmatrix}1&2\\3&4\end{pmatrix}}\right)}\right)}\cdot {\left({\begin{pmatrix}1&0\\0&1\end{pmatrix}}+{\left({\begin{pmatrix}3&-2\\0&5\end{pmatrix}}+{\begin{pmatrix}1&0\\0&1\end{pmatrix}}\right)}\right)}&={\begin{pmatrix}2&2\\3&5\end{pmatrix}}\cdot {\begin{pmatrix}5&-2\\0&7\end{pmatrix}}\\&={\begin{pmatrix}10&10\\15&29\end{pmatrix}}.\end{aligned}}$
Es ist
${}{\left({\left(5-{\left(-1\right)}\right)}-0\right)}-{\left({\left(-1\right)}-{\left(0-{\left(-1\right)}\right)}\right)}=6-{\left(-2\right)}=8\,.$
Es ist
${}{\begin{aligned}{\left({\left(\emptyset \cup \{1,2,3,4,5\}\right)}\cup \emptyset \right)}\cap {\left(\{1,2,3,4,5\}\cup {\left(\{2,4\}\cup \{1,2,3,4,5\}\right)}\right)}&=\{1,2,3,4,5\}\cap \{1,2,3,4,5\}\\&=\{1,2,3,4,5\}.\end{aligned}}$

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