{"id":28166,"date":"2021-04-09T13:00:38","date_gmt":"2021-04-09T13:00:38","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28166"},"modified":"2024-09-03T17:49:45","modified_gmt":"2024-09-03T17:49:45","slug":"provas-por-inducao-generalizacao-de-de-morgan-e-distribuicao","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/pt\/provas-por-inducao-generalizacao-de-de-morgan-e-distribuicao\/","title":{"rendered":"Provas por Indu\u00e7\u00e3o: Generaliza\u00e7\u00e3o de De Morgan e Distribui\u00e7\u00e3o"},"content":{"rendered":"<p><center><\/p>\n<h1>Provas por Indu\u00e7\u00e3o: Regras Generalizadas de De Morgan e Distribui\u00e7\u00e3o<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>RESUMO<\/strong><br \/><em>Esta aula aborda o tema das provas por indu\u00e7\u00e3o em matem\u00e1tica e l\u00f3gica proposicional. S\u00e3o explicados os dois tipos de provas existentes: as provas internas ou dedutivas, que se baseiam nas regras da l\u00f3gica, e as provas externas ou metamatem\u00e1ticas, que s\u00e3o necess\u00e1rias para provar afirma\u00e7\u00f5es que se referem \u00e0 pr\u00f3pria l\u00f3gica. A Indu\u00e7\u00e3o Matem\u00e1tica \u00e9 introduzida como um m\u00e9todo de demonstra\u00e7\u00e3o que permite provar que certas afirma\u00e7\u00f5es valem para todos os n\u00fameros naturais. Um exemplo \u00e9 apresentado com a demonstra\u00e7\u00e3o correspondente, e as formas generalizadas das leis de De Morgan e das leis distributivas em l\u00f3gica proposicional s\u00e3o explicadas, junto com suas respectivas demonstra\u00e7\u00f5es por indu\u00e7\u00e3o. Esta aula \u00e9 de grande import\u00e2ncia para entender os fundamentos da matem\u00e1tica e da l\u00f3gica e para aplic\u00e1-los em diferentes \u00e1reas do conhecimento.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>OBJETIVOS DE APRENDIZAGEM:<\/strong><br \/>\nAo final desta aula, o aluno ser\u00e1 capaz de:\n<\/p>\n<ol>\n<li><strong>Reconhecer<\/strong> os dois tipos de provas que devem ser distinguidos: provas internas ou dedutivas e provas externas ou metamatem\u00e1ticas.<\/li>\n<li><strong>Aplicar<\/strong> indu\u00e7\u00e3o matem\u00e1tica para fazer demonstra\u00e7\u00f5es sobre os n\u00fameros naturais e na l\u00f3gica proposicional.<\/li>\n<li><strong>Utilizar<\/strong> as nota\u00e7\u00f5es de conjun\u00e7\u00f5es e disjun\u00e7\u00f5es para escrever express\u00f5es da l\u00f3gica proposicional.<\/li>\n<li><strong>Compreender<\/strong> a forma generalizada das leis de De Morgan e de Distribui\u00e7\u00e3o na L\u00f3gica Proposicional.<\/li>\n<li><strong>Compreender<\/strong> o conceito de hip\u00f3tese de indu\u00e7\u00e3o e seu papel na demonstra\u00e7\u00e3o por indu\u00e7\u00e3o.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong>\u00cdNDICE<\/strong><br \/>\n<a href=\"#1\">PROVAS INTERNAS E EXTERNAS<\/a><br \/>\n<a href=\"#2\">PROVAS POR INDU\u00c7\u00c3O MATEM\u00c1TICA<\/a><br \/>\n<a href=\"#3\">PROVAS POR INDU\u00c7\u00c3O NA L\u00d3GICA PROPOSICIONAL<\/a><br \/>\n<a href=\"#4\">FORMA GENERALIZADA DAS LEIS DE DE MORGAN<\/a><br \/>\n<a href=\"#5\">FORMA GENERALIZADA DAS LEIS DISTRIBUTIVAS<\/a><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/eJQcNPrKyW0\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Provas internas e externas<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=212s\" target=\"_blank\" rel=\"noopener\"><strong>Existem dois tipos de provas que devem ser distinguidos.<\/strong><\/a> At\u00e9 agora, vimos muitos exemplos de provas formais. Esse tipo de prova emerge das regras da l\u00f3gica. Tais provas s\u00e3o ditas como ocorrendo \u00abdentro da l\u00f3gica\u00bb e, portanto, tamb\u00e9m s\u00e3o referidas como \u00abprovas internas\u00bb ou dedutivas. Essas provas formais t\u00eam um escopo limitado, pois s\u00f3 servem para provar afirma\u00e7\u00f5es que podem ser escritas na linguagem da l\u00f3gica. No entanto, podemos querer provar algumas coisas sobre a pr\u00f3pria l\u00f3gica. Podemos querer provar que todas as afirma\u00e7\u00f5es da l\u00f3gica proposicional satisfazem uma certa propriedade. Tais afirma\u00e7\u00f5es, que se referem \u00e0 pr\u00f3pria l\u00f3gica, n\u00e3o podem ser estabelecidas ou provadas dentro da l\u00f3gica. Para provar tais afirma\u00e7\u00f5es, utilizamos uma prova externa. As provas externas \u00e0s vezes s\u00e3o chamadas de \u00abmetamatem\u00e1ticas\u00bb e j\u00e1 nos deparamos com esse tipo de coisa, como quando vimos o (meta)teorema da dedu\u00e7\u00e3o. \u00c9 aqui que contextualizamos as provas indutivas.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Provas por Indu\u00e7\u00e3o Matem\u00e1tica<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=359s\" target=\"_blank\" rel=\"noopener\"><strong>A Indu\u00e7\u00e3o Matem\u00e1tica \u00e9 um m\u00e9todo de demonstra\u00e7\u00e3o<\/strong><\/a> que nos permite provar que algumas coisas valem para todos os n\u00fameros naturais.<\/p>\n<p style=\"text-align: justify;\"><strong>EXEMPLO:<\/strong> \u00c9 poss\u00edvel provar que qualquer n\u00famero da forma <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^n - 4^n<\/span><\/span>, onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> \u00e9 qualquer n\u00famero natural, \u00e9 sempre divis\u00edvel por 7.<br \/>\n<strong>DEMONSTRA\u00c7\u00c3O:<\/strong> Se observarmos o que ocorre com <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1<\/span><\/span>, veremos que:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^1 - 4^1 = 7<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">que obviamente \u00e9 divis\u00edvel por 7.<\/p>\n<p style=\"text-align: justify;\">Agora, suponha que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^n - 4^n<\/span><\/span> seja divis\u00edvel para um <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k.<\/span><\/span> A partir disso, provaremos que essa express\u00e3o tamb\u00e9m ser\u00e1 v\u00e1lida para <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k+1.<\/span><\/span> Podemos fazer isso da seguinte forma:<\/p>\n<table\">\n<tbody>\n<tr>\n<td style=\"text-align: right;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^{k+1} - 4^{k+1}<\/span><\/span><\/td>\n<td style=\"text-align: left;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">=11 \\cdot 11^{k} - 4 \\cdot 4^{k}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: right;\"><\/td>\n<td style=\"text-align: left;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">=11 \\cdot 11^{k} - (11-7) \\cdot 4^{k}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: right;\"><\/td>\n<td style=\"text-align: left;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">=11 \\cdot 11^{k} - 11 \\cdot 4^{k} + 7\\cdot 4^{k}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: right;\"><\/td>\n<td style=\"text-align: left;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">=11 ( 11^{k} - 4^{k} ) + 7\\cdot 4^{k}<\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\">Portanto, se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^k - 4^k<\/span><\/span> for divis\u00edvel por 7, consequentemente <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11 ( 11^{k} - 4^{k} ) + 7\\cdot 4^{k}<\/span><\/span> tamb\u00e9m ser\u00e1 divis\u00edvel por 7, o que \u00e9 o mesmo que dizer que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^{k+1} - 4^{k+1}<\/span><\/span> \u00e9 divis\u00edvel por 7. A partir disso, temos que se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^k - 4^k<\/span><\/span> for divis\u00edvel para <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=1<\/span><\/span>, ent\u00e3o ser\u00e1 para <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=2, k=3, k=4,\\cdots<\/span><\/span> e assim por diante, e, portanto, divis\u00edvel para qualquer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{N}.<\/span><\/span> Quando isso ocorre, dizemos que a indu\u00e7\u00e3o est\u00e1 completa. \u25a0<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Provas por Indu\u00e7\u00e3o na L\u00f3gica Proposicional<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=775s\" target=\"_blank\" rel=\"noopener\"><strong>Para as provas por indu\u00e7\u00e3o que realizaremos a seguir,<\/strong><\/a> ser\u00e1 necess\u00e1rio introduzir primeiro a seguinte conven\u00e7\u00e3o de nota\u00e7\u00e3o<\/p>\n<p style=\"text-align: justify;\"><strong>NOTA\u00c7\u00c3O:<\/strong> Sejam <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1,\\cdots, F_n<\/span><\/span> um conjunto finito de express\u00f5es quaisquer da l\u00f3gica proposicional. As conjun\u00e7\u00f5es e disjun\u00e7\u00f5es dessas express\u00f5es s\u00e3o introduzidas da seguinte maneira:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\bigwedge_{i=1}^n F_i := F_1\\wedge \\cdots \\wedge F_n<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\bigvee_{i=1}^n F_i := F_1\\vee \\cdots \\vee F_n<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Com isso, podemos agora lidar com as seguintes duas formas generalizadas.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Forma Generalizada das Leis de De Morgan<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=829s\" target=\"_blank\" rel=\"noopener\"><strong>Dado um conjunto finito de express\u00f5es da l\u00f3gica proposicional<\/strong><\/a> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1,\\cdots, F_n,<\/span><\/span>, sempre se cumprir\u00e3o as seguintes duas propriedades:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\neg\\left(\\bigwedge_{i=1}^n F_i \\right) \\equiv \\left( \\bigvee_{i=1}^n \\neg F_i \\right)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\neg\\left(\\bigvee_{i=1}^n F_i \\right) \\equiv \\left( \\bigwedge_{i=1}^n \\neg F_i \\right)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><strong>DEMONSTRA\u00c7\u00c3O:<\/strong> Primeiro, provaremos por indu\u00e7\u00e3o sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\neg\\left(\\bigwedge_{i=1}^n F_i \\right) \\equiv \\left( \\bigvee_{i=1}^n \\neg F_i \\right)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Primeiro, devemos verificar o que acontece com o caso inicial <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1.<\/span><\/span> Neste caso, \u00e9 claro que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\neg F_1 \\equiv \\neg\\left(\\bigwedge_{i=1}^1F_i\\right)\\equiv \\left(\\bigvee_{i=1}^n \\neg F_i \\right) \\equiv\\neg F_1<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Agora suponha que a propriedade funcione para algum <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k;<\/span><\/span> ou seja, dado um conjunto finito de express\u00f5es <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1, F_2, \\cdots, F_k<\/span><\/span>, temos que:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\neg\\left(\\bigwedge_{i=1}^k F_i\\right) \\equiv \\left(\\bigvee_{i=1}^k \\neg F_i\\right)<\/span><\/span><\/p>\n<p style=\"text-align: justify\">Ent\u00e3o, provaremos que, consequentemente, vale:<\/p>\n<p style=\"text-align: justify\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\neg\\left(\\bigwedge_{i=1}^{k+1} F_i\\right) \\equiv \\left(\\bigvee_{i=1}^{k+1} \\neg F_i\\right)<\/span><\/span><\/p>\n<p style=\"text-align: justify\">Usando a defini\u00e7\u00e3o de conjun\u00e7\u00e3o, temos que:<\/p>\n<p style=\"text-align: justify\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\neg\\left(\\bigwedge_{i=1}^{k+1} F_i\\right) := \\neg\\left[\\left(\\bigwedge_{i=1}^{k} F_i\\right) \\wedge F_{k+1}\\right]<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Sobre essa express\u00e3o, podemos aplicar as leis de De Morgan (a usual sobre dois termos) para obter:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\neg\\left(\\bigwedge_{i=1}^{k+1} F_i\\right)\\equiv \\left[\\neg\\left(\\bigwedge_{i=1}^{k} F_i\\right) \\vee \\neg F_{k+1}\\right]<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Agora, se aplicarmos a hip\u00f3tese de indu\u00e7\u00e3o, obteremos:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\neg\\left(\\bigwedge_{i=1}^{k+1} F_i\\right)\\equiv \\left[ \\left(\\bigvee_{i=1}^k \\neg F_i\\right) \\vee \\neg F_{k+1}\\right] := \\left(\\bigvee_{i=1}^{k+1}\\neg F_i \\right)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">E por esse motivo a indu\u00e7\u00e3o est\u00e1 completa e a propriedade vale para qualquer <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> em geral. A segunda rela\u00e7\u00e3o pode ser obtida de forma completamente an\u00e1loga, por isso deixarei como exerc\u00edcio para o leitor, muahahaha!<\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h2>Forma Generalizada das Leis Distributivas<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=1205s\" target=\"_blank\" rel=\"noopener\"><strong>De forma semelhante \u00e0s leis de De Morgan<\/strong><\/a>, as leis de distribui\u00e7\u00e3o podem ser generalizadas da seguinte maneira. Sejam <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{F_1, \\cdots, F_n\\}<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{G_1,\\cdots, G_m\\}<\/span><\/span> dois conjuntos finitos de express\u00f5es quaisquer, ent\u00e3o as seguintes equival\u00eancias s\u00e3o v\u00e1lidas:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^m G_j \\right) \\right] \\equiv \\left[\\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^m(F_i\\vee G_j) \\right) \\right]<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigvee_{i=1}^n F_i \\right) \\wedge \\left(\\bigvee_{j=1}^m G_j \\right) \\equiv \\left[\\bigvee_{i=1}^n\\left(\\bigvee_{j=1}^m(F_i\\wedge G_j) \\right) \\right]<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><strong>DEMONSTRA\u00c7\u00c3O:<\/strong> Para construir esta demonstra\u00e7\u00e3o, devemos realizar uma dupla indu\u00e7\u00e3o, sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> e <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m.<\/span><\/span> Em seguida, farei a indu\u00e7\u00e3o primeiro sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> e depois sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m<\/span><\/span> para a express\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^m G_j \\right) \\right] \\equiv \\left[\\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^m(F_i\\vee G_j) \\right) \\right]<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Se tomarmos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=1,<\/span><\/span> ent\u00e3o esta express\u00e3o fica escrita como<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^1 G_j \\right) \\equiv \\left[\\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^1(F_i\\vee G_j) \\right) \\right].<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">O que \u00e9 o mesmo que dizer:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee G_1 \\equiv \\left[\\bigwedge_{i=1}^n\\left( F_i\\vee G_1 \\right) \\right].<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Agora, provaremos esta express\u00e3o por indu\u00e7\u00e3o sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Se tomarmos <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1,<\/span><\/span> ent\u00e3o a express\u00e3o se reduz a<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1 \\vee G_1 \\equiv F_1 \\vee G_1.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">O que j\u00e1 sabemos que \u00e9 verdadeiro. Agora, suponha que se cumpra para algum <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k<\/span><\/span>; ou seja, a hip\u00f3tese de indu\u00e7\u00e3o ser\u00e1:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^k F_i \\right) \\vee G_1 \\equiv \\left[\\bigwedge_{i=1}^k\\left( F_i\\vee G_1 \\right) \\right].<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Ent\u00e3o, a partir disso, mostraremos que, consequentemente, se cumpre para <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k+1.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Pela defini\u00e7\u00e3o de conjun\u00e7\u00e3o, temos que:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[\\left(\\bigwedge_{i=1}^{k+1}F_i \\right) \\vee G_1 := \\left[\\left(\\left(\\bigwedge_{i=1}^{k}F_i \\right)\\wedge F_{k+1} \\right) \\vee G_1 <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Agora, usando a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vee<\/span><\/span>-distribui\u00e7\u00e3o, teremos que:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[\\left(\\bigwedge_{i=1}^{k+1}F_i \\right) \\vee G_1 equiv \\left[\\left(\\left(\\bigwedge_{i=1}^{k}F_i \\right)\\vee G_{1} \\right) \\wedge \\left(F_{k+1} \\vee G_1 \\right) <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">E, bem neste ponto, podemos usar a hip\u00f3tese de indu\u00e7\u00e3o para obter:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[\\left(\\bigwedge_{i=1}^{k+1}F_i \\right) \\vee G_1 equiv \\left[\\left(\\bigwedge_{i=1}^k\\left( F_i\\vee G_1 \\right) \\right) \\wedge \\left(F_{k+1} \\vee G_1 \\right) := \\left[\\bigwedge_{i=1}^{k+1}(F_{i}\\vee G_1 <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Portanto, provamos por indu\u00e7\u00e3o que para todo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{N}<\/span><\/span> se cumpre:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right)\\vee G_1 equival a \\left[\\bigwedge_{i=1}^n(F_i\\vee G_1)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Completando a indu\u00e7\u00e3o sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span>, verificamos que o caso inicial para <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=1,<\/span><\/span> funciona. Agora s\u00f3 precisamos completar a indu\u00e7\u00e3o sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m.<\/span><\/span> Para fazer isso, estabelecemos a hip\u00f3tese de indu\u00e7\u00e3o para um <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=l<\/span><\/span>, ou seja, funciona:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^l G_j equival a \\left[\\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^l(F_i\\vee G_j <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">e a partir disso, provaremos que tamb\u00e9m funciona para <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=l+1.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Come\u00e7ando, como sempre, pela defini\u00e7\u00e3o de conjun\u00e7\u00e3o, temos que<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^{l+1} G_j := \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\left(\\bigwedge_{j=1}^{l} G_j \\right) \\wedge G_{l+1]<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Agora, usando a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vee<\/span><\/span>-distribui\u00e7\u00e3o, teremos:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^{l+1} G_j equiv a \\left[ \\left( \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left( \\bigwedge_{j=1}^l G_j \\right) \\right) \\wedge \\left( \\left( \\bigwedge_{i=1}^n F_i \\right)\\vee G_{l+1 <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Consequentemente, usando a hip\u00f3tese de indu\u00e7\u00e3o, voc\u00ea pode escrever:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^{l+1} G_j equiv a \\left[ \\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^l(F_i\\vee G_j ) \\wedge \\left( \\left( \\bigwedge_{i=1}^n F_i \\right)\\vee G_{l+1 <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">E se agora tomarmos o resultado da indu\u00e7\u00e3o sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^{l+1} G_j equiv a \\left[ \\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^l(F_i\\vee G_j ) \\wedge \\left( \\bigwedge_{i=1}^n (F_i \\vee G_{l+1 )<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Que, finalmente, pela defini\u00e7\u00e3o de conjun\u00e7\u00e3o, temos:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^{l+1} G_j equiv a \\left[ \\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^{l+1}(F_i\\vee G_j )<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">E, portanto, a indu\u00e7\u00e3o sobre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m<\/span><\/span> est\u00e1 completa e a express\u00e3o:<\/p>\n<p style=\"text-align: justify;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\bigwedge_{j=1}^{m} G_j equiv a \\left[ \\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^{m}(F_i\\vee G_j <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">\u00e9 v\u00e1lida para todos os <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n,m\\in\\mathbb{N<\/span><\/span>.<\/p>\n<p style=\"text-align: justify;\">Essa explora\u00e7\u00e3o das provas por indu\u00e7\u00e3o demonstrou como t\u00e9cnicas rigorosas de demonstra\u00e7\u00e3o matem\u00e1tica podem ser aplicadas n\u00e3o apenas no \u00e2mbito dos n\u00fameros naturais, mas tamb\u00e9m na l\u00f3gica proposicional. Atrav\u00e9s da indu\u00e7\u00e3o, estabelecemos a validade das formas generalizadas das leis de De Morgan e das leis distributivas, refor\u00e7ando assim a compreens\u00e3o dos fundamentos l\u00f3gicos subjacentes a diversas \u00e1reas do conhecimento matem\u00e1tico. Essa abordagem n\u00e3o \u00e9 apenas essencial para o desenvolvimento de habilidades de racioc\u00ednio abstrato, mas tamb\u00e9m serve como uma ferramenta poderosa para enfrentar problemas complexos em matem\u00e1tica e al\u00e9m.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Provas por Indu\u00e7\u00e3o: Regras Generalizadas de De Morgan e Distribui\u00e7\u00e3o RESUMOEsta aula aborda o tema das provas por indu\u00e7\u00e3o em matem\u00e1tica e l\u00f3gica proposicional. S\u00e3o explicados os dois tipos de provas existentes: as provas internas ou dedutivas, que se baseiam nas regras da l\u00f3gica, e as provas externas ou metamatem\u00e1ticas, que s\u00e3o necess\u00e1rias para provar [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28154,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":12,"footnotes":""},"categories":[607,621,571],"tags":[],"class_list":["post-28166","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-logica-matematica-pt","category-logica-proposicional-pt","category-matematica-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Provas por Indu\u00e7\u00e3o: Generaliza\u00e7\u00e3o de De Morgan e Distribui\u00e7\u00e3o - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Tudo sobre como realizar Provas por Indu\u00e7\u00e3o em matem\u00e1tica e l\u00f3gica. 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