backup: 2026-04-17 11:28
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---
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status: seed
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type: concept
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tags:
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- math
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created: 2025-12-17
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updated: 2026-03-05
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title: Доказательство законов Де Моргана
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---
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# Доказательство законов Де Моргана
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## $\overline{A \cup B} = \overline{A} \cap \overline{B}$
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$$1 - \mu_{A \cup B}(u)=1 - \max (\mu_{A}(u);\ \mu_{B}(u))$$
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$$\overline{A} \cap \overline{B} = \min(1-\mu_{A}(u);\ 1-\mu_{B}(u))$$
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Предположим, что $\mu_{B}(u) > \mu_{A}(u)$, тогда $1-\mu_{B}(u)=1-\mu_{B}(u)$
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## $\overline{A \cap B} = \overline{A} \cup \overline{B}$
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$$1-\mu_{A \cap B}=1 -min(\mu_{A}(u);\ \mu_{B}(u))$$
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$$\overline{A} \cup \overline{B} = max(1 - \mu_{A}(u);\ 1-\mu_{B}(u))$$
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Пусть $\mu_{B}(u)>\mu_{A}(u)$, тогда $1-\mu_{A}(u)=1-\mu_{A}(u)$
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## $\overline{A * B} = \overline{A} + \overline{B}$
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$$(\mu_{\overline{A}}(u) + \mu_{\overline{B}}(u) - \mu_{\overline{A}}(u)\mu_{\overline{B}}(u)) = $$
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$$= 1-\mu_{A}(u) + 1 - \mu_{B}(u) - 1 + \mu_{A}(u) + \mu_{B}(u)-\mu_{A}(u)\mu_{B}(u)=$$
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$$= 1 - \mu_{A}(u) \mu_{B}(u)$$
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## $\overline{A + B} = \overline{A} * \overline{B}$
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$$(1-\mu_{A}(u))*(1-\mu_{B}(u)) = 1-\mu_{A}(u)-\mu_{B}(u)+\mu_{A}(u)\mu_{B}(u) = $$
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$$1 - (\mu_{A}(u)+\mu_{B}(u)-\mu_{A}(u)\mu_{B}(u))$$
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