{"id":169223,"date":"2023-02-10T14:29:15","date_gmt":"2023-02-10T20:29:15","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=169223"},"modified":"2025-12-11T12:05:35","modified_gmt":"2025-12-11T18:05:35","slug":"de-morgans-laws","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/de-morgans-laws\/","title":{"rendered":"De Morgan&#8217;s Laws"},"content":{"rendered":"<h1>De Morgan&#8217;s Laws<\/h1>\n\n\t\t\t<div id=\"mmDeferVideoEncompass_GjETVVN945I\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_GjETVVN945I\" data-source-videoID=\"GjETVVN945I\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"De Morgan&#8217;s Laws Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"De Morgan&#8217;s Laws\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_GjETVVN945I:hover {cursor:pointer;} img#videoThumbnailImage_GjETVVN945I {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1794-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_GjETVVN945I\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_GjETVVN945I\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"De Morgan&#8217;s Laws\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_GjETVVN945I\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_GjETVVN945I\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_GjETVVN945I\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Up to this point, we\u2019ve done some introductory work with <a href=\"https:\/\/www.mometrix.com\/academy\/set-operations-with-venn-diagrams\/\" style=\"text-decoration: underline;\"><strong>Venn diagrams<\/strong><\/a>, <a href=\"https:\/\/www.mometrix.com\/academy\/intro-to-set-theory\/\" style=\"text-decoration: underline;\"><strong>set theory<\/strong><\/a>, unions, and intersections. How do we handle more complex set theory expressions, like the complement of a union or the complement of an intersection?<\/p>\n<div class=\"examplesentence\">\\((A\\cap B)&#8217;\\)<br \/>\n&nbsp;<br \/>\n\\((A\\cap B)&#8217;\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>De Morgan\u2019s laws are helpful tools for representing complex expressions like these in more understandable ways.<\/p>\n<p>To understand De Morgan\u2019s laws, let\u2019s temporarily replace the complement sign with the word \u201cnot.\u201d Starting with the complement of \\(A\\cap B\\), we get:<\/p>\n<div class=\"examplesentence\">Not \\((A\\cap B)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Using a Venn diagram, we can see this as everything <em>except<\/em> the intersection of \\(A\\) and \\(B\\).<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-169520\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-1.jpg\" alt=\"Venn diagram seen as everything except the intersection of A and B.\" width=\"536\" height=\"300\"style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-1.jpg 536w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-1-300x168.jpg 300w\" sizes=\"auto, (max-width: 536px) 100vw, 536px\" \/><\/p>\n<p>One of De Morgan\u2019s law says that we can distribute this \u201cnot\u201d to the sets inside the parentheses \\(IF\\) we switch that intersection to a union. So \u201cnot \\((A\\cap B)\\)\u201d becomes \u201cnot \\(A\\) union not \\(B\\).\u201d<\/p>\n<div class=\"examplesentence\">Not \\((A\\cap B)\\rightarrow  A&#8217;\\cup B&#8217;\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s graph the complements of \\(A\\) and \\(B\\) to verify that their union is the same as the complement of \\((A\\cap B)\\).<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-169517\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-2.jpg\" alt=\"Venn graph of complements A and B\" width=\"652\" height=\"214\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-2.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-2-300x98.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>The only thing not covered by \\(A&#8217;\\) or by \\(B&#8217;\\) is the intersection of \\(A\\) and \\(B\\). So, the union of \\(A&#8217;\\) and \\(B&#8217;\\) covers everything except for that small area. This is exactly what we saw with the complement of \\(A\\cap B\\) so, we have verified De Morgan\u2019s law that \\((A\\cap B&#8217;)=A&#8217;\\cup B&#8217;\\).<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-169514\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-3.jpg\" alt=\"Venn graph of intersection A and B\" width=\"652\" height=\"363\" style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-3.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-3-300x167.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<div class=\"examplesentence\">Not \\((A \\cap B)=\\) Not \\(A\\text{ } \\cup\\) Not \\(B\\)<br \/>\n&nbsp;<br \/>\n\\((A\\cap B)&#8217;=A&#8217;\\cup B&#8217;\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s look at the next statement of De Morgan\u2019s law. Consider the complement of \\(A\\) <em>union<\/em> \\(B\\). We know that \\(A\\cup B\\) is everything in \\(A\\) with everything in \\(B\\), so the complement of this union is everything that\u2019s outside of \\(A\\) and outside of \\(B\\).<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-169511\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-4.jpg\" alt=\"Venn diagram of a complement that's outside A and B\" width=\"652\" height=\"363\" style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-4.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-4-300x167.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>As with the previous statement of De Morgan\u2019s law, we can rewrite \\((A\\cup B)&#8217;\\) as \u201cnot\u201d \\(A\\cup B\\).<\/p>\n<div class=\"examplesentence\">Not \\((A\\cup B)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Then we can distribute the \u201cnot\u201d by changing the union to an intersection.<\/p>\n<div class=\"examplesentence\">Not \\((A\\cup B)\\rightarrow\\) Not \\(A\\text{ } \\cap\\) Not \\(B\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>So, the complement of \\((A\\cup B\\) is equal to \\(A&#8217;\\cap B&#8217;\\). Let\u2019s check this with a Venn diagram. Once again, \\(A&#8217;\\) is everything outside of \\(A\\), and \\(B&#8217;\\) is everything outside of \\(B\\).<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-169508\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-5.jpg\" alt=\"Venn diagram of everything outside of A and B.\" width=\"652\" height=\"237\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-5.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-5-300x109.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>The intersection of these areas is only what is counted in both of them, which is the area completely outside of \\(A\\) and \\(B\\). As we said a moment ago, this area can also be called \\((A\\cup B)&#8217;\\).<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-169505\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-6.jpg\" alt=\"Venn Diagram\" width=\"652\" height=\"363\" style=\"box-shadow: 1.5px 1.5px 5px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-6.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/De-Morgans-Laws-6-300x167.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<div class=\"examplesentence\">\\((A\\cup B)&#8217;=A&#8217;\\cap B&#8217;\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>By this we have verified De Morgan\u2019s law that the complement of \\(A\\cup B\\) is equal to \\(A&#8217;\\cap B&#8217;\\).<\/p>\n<p>Let\u2019s now apply this knowledge to a couple example problems. <\/p>\n<p>What is another way to write the complement of \\(A\\cap B&#8217;\\)?<\/p>\n<div class=\"examplesentence\">\\((A\\cap B&#8217;)^{&#8216;}=?\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>This example is a little different because we now have \\(B\\)\u2019s complement instead of \\(B\\). We can rewrite this using the same techniques as before. Let\u2019s start by writing the word \u201cnot\u201d instead of the complement symbol: Not \\((A \\cap B&#8217;)\\). Now, De Morgan\u2019s law says that we can distribute this \u201cnot\u201d to the \\(A\\) and \\(B&#8217;\\) inside the parentheses IF we change the intersection to a union:<\/p>\n<div class=\"examplesentence\">\\(\\rightarrow\\) Not \\(A\\text{ }\\cup \\) Not \\(B&#8217;\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>\u201cNot \\(B&#8217;\\)\u201d can be rewritten as simply \\(B\\), because the \u201cnot\u201d and the complement cancel each other out.<\/p>\n<div class=\"examplesentence\">\\(\\rightarrow\\) Not \\(A\\cup B\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s finish by rewriting \u201cNot \\(A\\)\u201d as \\(A&#8217;\\).<\/p>\n<div class=\"examplesentence\">\\(\\rightarrow A&#8217;\\cup B\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>This is as far as we can simplify this expression. So, the complement of \\(A\\cap B&#8217;\\) is equal to \\(A&#8217;\\) union \\(B\\).<\/p>\n<div class=\"examplesentence\">\\((A \\cap B&#8217;)&#8217;=A&#8217; \\cup B\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s finish with a word problem. Rewrite the following statement using De Morgan\u2019s laws: \u201cIt is not true that we are going to eat hamburgers or that we are not going to bake cookies.\u201d<\/p>\n<p>This statement is a bit confusing as it is currently written, but we have the tools to rewrite it in a clearer way. What can we already determine about this statement? Well, the phrase \u201cit is not true that\u201d indicates that the remainder of the sentence is to be negated. So, we can write this in a sort of set notation as the complement of (hamburgers or not cookies). <\/p>\n<div class=\"examplesentence\">(Hamburgers or Not cookies)&#8217;<\/div>\n<p>\n&nbsp;<\/p>\n<p>You\u2019ll recall that the word \u201cor\u201d in set theory is associated with the union of two sets. Let\u2019s write the union symbol in place of \u201cor.\u201d<\/p>\n<div class=\"examplesentence\">(Hamburgers \\(\\cup\\) Not cookies)&#8217;<\/div>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s now distribute the outer \u201cnot\u201d to hamburgers and \u201cnot cookies,\u201d remembering to switch the union to an intersection.<\/p>\n<div class=\"examplesentence\">Not hamburgers \\(\\cap\\) Not not cookies<\/div>\n<p>\n&nbsp;<\/p>\n<p>Here, the two nots before cookies form a double negative and cancel each other out, leaving us with:<\/p>\n<div class=\"examplesentence\">Not hamburgers \\(\\cap\\) Cookies<\/div>\n<p>\n&nbsp;<\/p>\n<p>So, if we say, \u201cIt is not true that we are going to eat hamburgers or that we are not going to bake cookies,\u201d we mean that \u201cwe are not going to eat hamburgers <em>and<\/em> we are going to bake cookies.\u201d A more natural way of saying this is, \u201cwe are not going to eat hamburgers, <em>but<\/em> we are going to bake cookies.\u201d The word \u201cbut\u201d can be used in place of the intersection\u2019s \u201cand\u201d whenever one of the sets is negated.<\/p>\n<p>As we have seen, De Morgan\u2019s laws state that a complement can be distributed to the sets inside parentheses as long as the union inside is swapped for an intersection, or if it is an intersection, swap it for a union when distributing the complement. <\/p>\n<div class=\"examplesentence\">Not \\((A \\cup B\\rightarrow\\) Not \\(A\\text{ } \\cap\\) Not \\(B\\)<br \/>\n&nbsp;<br \/>\nNot \\((A\\cap B)\\rightarrow \\)Not \\(A \\text{ }\\cap\\) Not \\( B\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>De Morgan\u2019s laws are simple yet effective at turning complicated set theory expressions into more manageable ones.<\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<\/div>\n\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/discrete-math\/\"><strong>Return to Discrete Math Videos<\/strong><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>De Morgan&#8217;s Laws Return to Discrete Math Videos<\/p>\n","protected":false},"author":22,"featured_media":169229,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-169223","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-discrete-math-videos","7":"page_category-video-pages-for-study-course-sidebar-ad","8":"page_type-video"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/169223","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=169223"}],"version-history":[{"count":4,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/169223\/revisions"}],"predecessor-version":[{"id":184331,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/169223\/revisions\/184331"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/169229"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=169223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}