[nicolas-alan]

\setminus&space;C&space;=&space;\left&space;(&space;A&space;\setminus&space;C&space;\right&space\setminus&space;\left&space;(&space;B&space;\setminus&space;C&space;\right&space&space;\Leftrightarrow&space;\begin{Bmatrix}&space;(a):\left&space;(&space;A&space;\setminus&space;B&space;\right&space\setminus&space;C&space;\subseteq&space;\left&space;(&space;A&space;\setminus&space;C&space;\right&space\setminus&space;\left&space;(&space;B&space;\setminus&space;C&space;\right&space&space;\\&space;(b):\left&space;(&space;A&space;\setminus&space;C&space;\right&space\setminus&space;\left&space;(&space;B&space;\setminus&space;C&space;\right&space&space;\subseteq&space;\left&space;(&space;A&space;\setminus&space;B&space;\right&space\setminus&space;C&space;\end{Bmatrix}" title="\left ( A \setminus B \right )\setminus C = \left ( A \setminus C \right )\setminus \left ( B \setminus C \right ) \Leftrightarrow \begin{Bmatrix} (a):\left ( A \setminus B \right )\setminus C \subseteq \left ( A \setminus C \right )\setminus \left ( B \setminus C \right ) \\ (b):\left ( A \setminus C \right )\setminus \left ( B \setminus C \right ) \subseteq \left ( A \setminus B \right )\setminus C \end{Bmatrix}">

(b) доказал:

\setminus&space;\left&space;(&space;B&space;\setminus&space;C&space;\right)\overset{1}&space;{\rightarrow}&space;\\&space;\overset{1}&space;{\rightarrow}&space;\left&space;[(x&space;\in&space;A&space\wedge&space;(x&space;\notin&space;C&space&space;\right&space;]\wedge&space;x&space;\notin&space;\left&space;(&space;B&space;\setminus&space;C&space;\right)&space;\Rightarrow&space;\\&space;\Rightarrow&space;\left&space;[(x&space;\in&space;A&space\wedge&space;(x&space;\notin&space;C&space&space;\right&space;]\wedge&space;x&space;\in&space;\overline{&space;\left&space;(&space;B&space;\setminus&space;C&space;\right)&space;}&space;\Rightarrow&space;\\&space;\Rightarrow&space;\left&space;[(x&space;\in&space;A&space\wedge&space;(x&space;\notin&space;C&space&space;\right&space;]\wedge\overline{&space;\left&space;[(x&space;\in&space;B&space\wedge&space;(x&space;\notin&space;C&space&space;\right&space;]}\overset{2}&space;{\rightarrow}&space;\\&space;\overset{2}&space;{\rightarrow}&space;\left&space;[(x&space;\in&space;A&space\wedge&space;(x&space;\notin&space;C&space&space;\right&space;]\wedge&space;\left&space;[(x&space;\notin&space;B&space&space;\vee&space;(x&space;\in&space;C&space&space;\right&space;]&space;\overset{3}&space;{\rightarrow}&space;\\&space;\overset{3}&space;{\rightarrow}&space;\left&space;\langle&space;\left&space;[(x&space;\in&space;A&space\wedge&space;(x&space;\notin&space;C&space&space;\right&space;]&space;\wedge&space;(x&space;\notin&space;B&space&space;\right&space;\rangle&space;\vee&space;\left&space;\langle&space;\left&space;[(x&space;\in&space;A&space\wedge&space;(x&space;\notin&space;C&space&space;\right&space;]&space;\wedge&space;(x&space;\in&space;C&space&space;\right&space;\rangle&space;\overset{4}&space;{\rightarrow}" title="x \in \left ( A \setminus C \right )\setminus \left ( B \setminus C \right)\overset{1} {\rightarrow} \\ \overset{1} {\rightarrow} \left [(x \in A )\wedge (x \notin C ) \right ]\wedge x \notin \left ( B \setminus C \right) \Rightarrow \\ \Rightarrow \left [(x \in A )\wedge (x \notin C ) \right ]\wedge x \in \overline{ \left ( B \setminus C \right) } \Rightarrow \\ \Rightarrow \left [(x \in A )\wedge (x \notin C ) \right ]\wedge\overline{ \left [(x \in B )\wedge (x \notin C ) \right ]}\overset{2} {\rightarrow} \\ \overset{2} {\rightarrow} \left [(x \in A )\wedge (x \notin C ) \right ]\wedge \left [(x \notin B ) \vee (x \in C ) \right ] \overset{3} {\rightarrow} \\ \overset{3} {\rightarrow} \left \langle \left [(x \in A )\wedge (x \notin C ) \right ] \wedge (x \notin B ) \right \rangle \vee \left \langle \left [(x \in A )\wedge (x \notin C ) \right ] \wedge (x \in C ) \right \rangle \overset{4} {\rightarrow}">

\wedge&space;\left&space;[(x&space;\notin&space;C&space&space;\wedge&space;(x&space;\notin&space;B&space&space;\right&space;]\right&space;\rangle&space;\vee&space;\langle&space;(x&space;\in&space;A&space\wedge&space;\underset{false}{\underbrace{&space;\left&space;[(x&space;\notin&space;C&space&space;\wedge&space;(x&space;\in&space;C&space&space;\right&space;]&space;\right}&space;}&space;\rangle&space;\overset{5}&space;{\rightarrow}&space;\\&space;\overset{5}&space;{\rightarrow}&space;\left&space;\langle&space;(x&space;\in&space;A&space\wedge&space;\left&space;[(x&space;\notin&space;C&space&space;\wedge&space;(x&space;\notin&space;B&space&space;\right&space;]&space;\right&space;\rangle&space;\vee&space;\underset{false}{\underbrace{&space;\left&space;[&space;(x&space;\in&space;A&space\wedge&space;(false)&space;\right&space;]}&space;}&space;\overset{6}&space;{\rightarrow}&space;\\&space;\overset{6}&space;{\rightarrow}&space;(x&space;\in&space;A&space\wedge&space;\left&space;[(x&space;\notin&space;C&space&space;\wedge&space;(x&space;\notin&space;B&space&space;\right&space;]&space;\overset{7}&space;{\rightarrow}&space;\\&space;\overset{7}&space;{\rightarrow}&space;(x&space;\in&space;A&space\wedge&space;\left&space;[(x&space;\notin&space;B&space&space;\wedge&space;(x&space;\notin&space;C&space&space;\right&space;]&space;\overset{4}&space;{\rightarrow}&space;\\&space;\overset{4}&space;{\rightarrow}&space;\left&space;[&space;(x&space;\in&space;A&space\wedge&space;(x&space;\notin&space;B&space&space;\right&space;]&space;\wedge&space;(x&space;\notin&space;C&space&space;\overset{1}&space;{\rightarrow}&space;\\&space;\overset{1}&space;{\rightarrow}&space;x&space;\in&space;\left&space;(&space;A&space;\setminus&space;B\right&space&space;\wedge&space;x&space;\notin&space;C&space;\overset{1}&space;{\rightarrow}" title="\overset{4,5} {\rightarrow} \left \langle (x \in A )\wedge \left [(x \notin C ) \wedge (x \notin B ) \right ]\right \rangle \vee \langle (x \in A )\wedge \underset{false}{\underbrace{ \left [(x \notin C ) \wedge (x \in C ) \right ] \right} } \rangle \overset{5} {\rightarrow} \\ \overset{5} {\rightarrow} \left \langle (x \in A )\wedge \left [(x \notin C ) \wedge (x \notin B ) \right ] \right \rangle \vee \underset{false}{\underbrace{ \left [ (x \in A )\wedge (false) \right ]} } \overset{6} {\rightarrow} \\ \overset{6} {\rightarrow} (x \in A )\wedge \left [(x \notin C ) \wedge (x \notin B ) \right ] \overset{7} {\rightarrow} \\ \overset{7} {\rightarrow} (x \in A )\wedge \left [(x \notin B ) \wedge (x \notin C ) \right ] \overset{4} {\rightarrow} \\ \overset{4} {\rightarrow} \left [ (x \in A )\wedge (x \notin B ) \right ] \wedge (x \notin C ) \overset{1} {\rightarrow} \\ \overset{1} {\rightarrow} x \in \left ( A \setminus B\right ) \wedge x \notin C \overset{1} {\rightarrow}">

&space;\wedge&space;x&space;\notin&space;C&space;\overset{1}&space;{\rightarrow}&space;\\&space;\overset{1}&space;{\rightarrow}&space;x&space;\in&space;\left&space;(&space;A&space;\setminus&space;B\right&space&space;\setminus&space;C&space;\overset{8}&space;{\rightarrow}&space;\\&space;\overset{8}&space;{\rightarrow}&space;\left&space;(&space;A&space;\setminus&space;B&space;\right&space&space;\setminus&space;\left&space;(&space;A&space;\setminus&space;B\right&space&space;\subseteq&space;\left&space;(&space;A&space;\setminus&space;B\right&space&space;\setminus&space;C" title="\overset{1} {\rightarrow} x \in \left ( A \setminus B\right ) \wedge x \notin C \overset{1} {\rightarrow} \\ \overset{1} {\rightarrow} x \in \left ( A \setminus B\right ) \setminus C \overset{8} {\rightarrow} \\ \overset{8} {\rightarrow} \left ( A \setminus B \right ) \setminus \left ( A \setminus B\right ) \subseteq \left ( A \setminus B\right ) \setminus C">

где

&space;\\&space;3&space;-&space;Morgano&space;\,&space;desnis&space;\,&space;-&space;\overline{&space;x&space;\wedge&space;y&space;}&space;=&space;\overline{x}&space;\vee&space;\overline{y}&space;\\&space;4&space;-&space;Asociatyvumas&space;\,&space;-&space;x&space;\wedge&space;(&space;y&space;\wedge&space;z&space&space;=&space;(&space;x&space;\wedge&space;y&space&space;\wedge&space;z&space;\\&space;5&space;-&space;Prestaros&space;\,&space;desnis&space;\,&space;-&space;x&space;\wedge&space;\overline{x}&space;=&space;0&space;\\&space;6&space;-&space;Konstantos&space;\,&space;savybe&space;\,&space;-&space;x&space;\vee&space;0&space;=&space;x&space;\\&space;7&space;-&space;Komutatyvumas&space;\,&space;x&space;\wedge&space;y&space;=&space;y&space;\wedge&space;x&space;\\&space;8&space;-&space;Poaibio&space;\,&space;apibrezimas" title="\\ 1- aibes\, skirtumo \, apibrezimas \\ 2 - Konjunkciojos \, distributyvumas \, disjunkcijos \, atzvilgiu \, \\ x \wedge( y \vee z) = ( x \wedge y) \vee (x \wedge z ) \\ 3 - Morgano \, desnis \, - \overline{ x \wedge y } = \overline{x} \vee \overline{y} \\ 4 - Asociatyvumas \, - x \wedge ( y \wedge z ) = ( x \wedge y ) \wedge z \\ 5 - Prestaros \, desnis \, - x \wedge \overline{x} = 0 \\ 6 - Konstantos \, savybe \, - x \vee 0 = x \\ 7 - Komutatyvumas \, x \wedge y = y \wedge x \\ 8 - Poaibio \, apibrezimas">

а как доказать первую часть - пока не знаю. можно конечно, всё шиворот наыворот сделать, прибавить 0 и потом расписать, но в math.stackexchange.com/questions/239875/element... сделано по-другому.

пойду в диари математиков спрашивать.