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  • diffhist De Morgan's laws 15:06 +1Kaiphom talk contribs(→‎Part 2: Though 'or' provides a correct implication, using this argument cannot conclusively claim that <math>x \not\in \overline{A} \cup \overline{B}</math>. 'and' is correct and gives a much stricter condition, allowing <math>x \not\in \overline{A} \cup \overline{B}</math> to be guaranteed. <math>x \in A</math> and <math>x \in B</math>, and thus <math>x \not\in \overline{A}</math> or <math>x \not\in \overline{B}</math>. However, that means <math>x \not\in \overline{A} \cup \overline{B...) Tags: Mobile edit Mobile web edit