Logica matematica/Insiemi: differenze tra le versioni
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Riga 125:
#<math>S_1 \cap S_2=\{x|x \in S_1\ e\ x \in S_2\} \subseteq \{x|x \in S_1\}=S_1</math>;
#analoga alla precedente.
==== Proprietà distributive ====
<math>\cup</math> e <math>\cap</math> sono legate dalle seguenti proprietà:
# <math>S_1 \cap (S_2 \cup S_3)=(S_1 \cap S_2) \cup (S_1 \cap S_3)</math>;
# <math>S_1 \cup (S_2 \cap S_3)=(S_1 \cup S_2) \cap (S_1 \cup S_3)</math>.
''Dimostrazione''<blockquote>1. <math>S_1 \cap (S_2 \cup S_3)</math></blockquote><blockquote><math>=\{x|x \in S_1\ e\ x \in (S_2 \cup S_3)\}</math></blockquote><blockquote><math>=\{x|x \in S_1\ e\ (x \in S_2\ oppure\ x \in S_3)\}</math></blockquote><blockquote><math>=\{x|(x \in S_1\ e\ x \in S_2)\ oppure\ (x \in S_1\ e\ x \in S_3)\}</math></blockquote><blockquote><math>=\{x|x \in (S_1 \cap S_2)\ oppure\ x \in (S_1 \cap S_3)\}</math></blockquote><blockquote><math>=(S_1 \cap S_2) \cup (S_1 \cap S_3)</math>;</blockquote><blockquote>2. <math>S_1 \cup (S_2 \cap S_3)</math></blockquote><blockquote><math>=\{x|x \in S_1\ oppure\ x \in (S_2 \cap S_3)\}</math></blockquote><blockquote><math>=\{x|x \in S_1\ oppure\ (x \in S_2\ e\ x \in S_3)\}</math></blockquote><blockquote><math>=\{x|(x \in S_1\ oppure\ x \in S_2)\ e\ (x \in S_1\ oppure\ x \in S_3)\}</math></blockquote><blockquote><math>=\{x|x \in (S_1 \cup S_2)\ e\ x \in (S_1 \cup S_3)\}</math></blockquote><blockquote><math>=(S_1 \cup S_2) \cap (S_1 \cup S_3)</math>.</blockquote>
== Insieme potenza ==
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