{"id":28272,"date":"2021-04-09T13:00:03","date_gmt":"2021-04-09T13:00:03","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28272"},"modified":"2024-09-03T22:10:45","modified_gmt":"2024-09-03T22:10:45","slug":"preuves-par-induction-generalisation-de-de-morgan-et-distribution","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/fr\/preuves-par-induction-generalisation-de-de-morgan-et-distribution\/","title":{"rendered":"Preuves par induction : G\u00e9n\u00e9ralisation de De Morgan et Distribution"},"content":{"rendered":"<p><center><\/p>\n<h1>Preuves par induction : R\u00e8gles g\u00e9n\u00e9ralis\u00e9es de De Morgan et Distribution<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>R\u00c9SUM\u00c9<\/strong><br \/><em>Dans ce cours, nous abordons le sujet des preuves par induction en math\u00e9matiques et en logique propositionnelle. Nous expliquons les deux types de preuves existants : les preuves internes ou d\u00e9ductives, qui reposent sur les r\u00e8gles de la logique, et les preuves externes ou m\u00e9tamath\u00e9matiques, qui sont n\u00e9cessaires pour prouver des affirmations concernant la logique elle-m\u00eame. Nous introduisons l&#8217;induction math\u00e9matique comme m\u00e9thode de d\u00e9monstration permettant de prouver que certaines affirmations sont valables pour tous les nombres naturels. Nous pr\u00e9sentons un exemple avec la d\u00e9monstration correspondante et expliquons les formes g\u00e9n\u00e9ralis\u00e9es des lois de De Morgan et des lois distributives en logique propositionnelle, ainsi que leurs d\u00e9monstrations par induction respectives. Ce cours est d&#8217;une grande importance pour comprendre les fondements des math\u00e9matiques et de la logique, et pour les appliquer dans divers domaines du savoir.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>OBJECTIFS D&#8217;APPRENTISSAGE :<\/strong><br \/>\n\u00c0 la fin de ce cours, l&#8217;\u00e9tudiant sera capable de :\n<\/p>\n<ol>\n<li><strong>Reconna\u00eetre<\/strong> les deux types de preuves \u00e0 distinguer : preuves internes ou d\u00e9ductives et preuves externes ou m\u00e9tamath\u00e9matiques.<\/li>\n<li><strong>Appliquer<\/strong> l&#8217;induction math\u00e9matique pour faire des d\u00e9monstrations sur les nombres naturels et en logique propositionnelle.<\/li>\n<li><strong>Utiliser<\/strong> les notations de conjonctions et de disjonctions pour \u00e9crire des expressions de la logique propositionnelle.<\/li>\n<li><strong>Comprendre<\/strong> la forme g\u00e9n\u00e9ralis\u00e9e des lois de De Morgan et des lois distributives en logique propositionnelle.<\/li>\n<li><strong>Comprendre<\/strong> le concept d&#8217;hypoth\u00e8se d&#8217;induction et son r\u00f4le dans la d\u00e9monstration par induction.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong>INDEX<\/strong><br \/>\n<a href=\"#1\">PREUVES INTERNES ET EXTERNES<\/a><br \/>\n<a href=\"#2\">PREUVES PAR INDUCTION MATH\u00c9MATIQUE<\/a><br \/>\n<a href=\"#3\">PREUVES PAR INDUCTION EN LOGIQUE PROPOSITIONNELLE<\/a><br \/>\n<a href=\"#4\">FORME G\u00c9N\u00c9RALIS\u00c9E DES LOIS DE DE MORGAN<\/a><br \/>\n<a href=\"#5\">FORME G\u00c9N\u00c9RALIS\u00c9E DES LOIS DISTRIBUTIVES<\/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>Preuves internes et externes<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=212s\" target=\"_blank\" rel=\"noopener\"><strong>Il existe deux types de preuves qu&#8217;il faut distinguer.<\/strong><\/a> Jusqu&#8217;\u00e0 pr\u00e9sent, nous avons vu de nombreux exemples de preuves formelles. Ce type de preuves \u00e9merge des r\u00e8gles de la logique. De telles preuves sont dites avoir lieu \u00ab\u00e0 l&#8217;int\u00e9rieur de la logique\u00bb, et nous nous y r\u00e9f\u00e9rons donc \u00e9galement sous le nom de \u00abpreuves internes\u00bb ou d\u00e9ductives. Ce type de preuves formelles a une port\u00e9e limit\u00e9e, car elles ne servent qu&#8217;\u00e0 prouver des affirmations qui peuvent \u00eatre \u00e9crites dans le langage de la logique. Cependant, nous pourrions vouloir prouver certaines choses concernant la logique elle-m\u00eame. Nous pourrions vouloir prouver que toutes les affirmations de la logique propositionnelle satisfont \u00e0 une certaine propri\u00e9t\u00e9. De telles affirmations qui se r\u00e9f\u00e8rent \u00e0 la logique elle-m\u00eame ne peuvent \u00eatre \u00e9tablies ni prouv\u00e9es \u00e0 l&#8217;int\u00e9rieur de la logique. Pour prouver de telles affirmations, nous utilisons une preuve externe. Les preuves externes sont parfois appel\u00e9es \u00abm\u00e9tamath\u00e9matiques\u00bb, et nous avons d\u00e9j\u00e0 rencontr\u00e9 ce type de choses, comme lorsque nous avons vu le (m\u00e9ta)th\u00e9or\u00e8me de d\u00e9duction. C&#8217;est ici que nous contextualisons les preuves inductives.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Preuves par Induction Math\u00e9matique<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=359s\" target=\"_blank\" rel=\"noopener\"><strong>L&#8217;induction math\u00e9matique est une m\u00e9thode de d\u00e9monstration<\/strong><\/a> qui nous permet de prouver que certaines choses sont valables pour tous les nombres naturels.<\/p>\n<p style=\"text-align: justify;\"><strong>EXEMPLE :<\/strong><br \/>\nIl est possible de prouver que tout nombre de la forme <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^n - 4^n<\/span><\/span>, o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> est un nombre naturel quelconque, est toujours divisible par 7.<br \/>\n<strong>D\u00c9MONSTRATION :<\/strong> Si nous observons ce qui se passe avec <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1<\/span><\/span>, nous verrons 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;\">ce qui, \u00e9videmment, est divisible par 7.<\/p>\n<p style=\"text-align: justify;\">Supposons maintenant que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^n - 4^n<\/span><\/span> soit divisible pour un <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k.<\/span><\/span> \u00c0 partir de cela, nous prouverons en cons\u00e9quence que cette expression sera \u00e9galement vraie pour <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k+1.<\/span><\/span> Nous pouvons le faire de la mani\u00e8re suivante :<\/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;\">Par cons\u00e9quent, si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^k - 4^k<\/span><\/span> est divisible par 7, en cons\u00e9quence <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11 ( 11^{k} - 4^{k} ) + 7\\cdot 4^{k}<\/span><\/span> le sera aussi, ce qui revient \u00e0 dire que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^{k+1} - 4^{k+1}<\/span><\/span> est divisible par 7. De l\u00e0, nous avons que si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^k - 4^k<\/span><\/span> est divisible pour <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=1<\/span><\/span>, alors il le sera pour <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=2, k=3, k=4,\\cdots<\/span><\/span> et ainsi de suite, et donc divisible pour tout <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{N}.<\/span><\/span> Lorsque cela se produit, on dit que l&#8217;induction est compl\u00e8te. \u25a0<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Preuves par Induction en Logique Propositionnelle<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=775s\" target=\"_blank\" rel=\"noopener\"><strong>Pour les preuves par induction que nous allons r\u00e9aliser ci-dessous,<\/strong><\/a> il sera d&#8217;abord n\u00e9cessaire d&#8217;introduire la convention de notation suivante<\/p>\n<p style=\"text-align: justify;\"><strong>NOTATION : <\/strong> Soient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1,\\cdots, F_n<\/span><\/span> un ensemble fini d&#8217;expressions quelconques de la logique propositionnelle. Les conjonctions et les disjonctions de ces expressions sont introduites par :<\/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;\">Avec cela, nous pourrons maintenant traiter les deux formes g\u00e9n\u00e9ralis\u00e9es suivantes.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Forme G\u00e9n\u00e9ralis\u00e9e des Lois 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>\u00c9tant donn\u00e9 un ensemble fini d&#8217;expressions de la logique propositionnelle<\/strong><\/a> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1,\\cdots, F_n,<\/span><\/span> les deux propri\u00e9t\u00e9s suivantes seront toujours v\u00e9rifi\u00e9es :<\/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>D\u00c9MONSTRATION :<\/strong> Nous prouverons d&#8217;abord par induction sur <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;\">Nous devons d&#8217;abord examiner ce qui se passe avec le cas initial <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1.<\/span><\/span> Dans ce cas, il est clair 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;\">Supposons maintenant que la propri\u00e9t\u00e9 fonctionne pour un certain <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k;<\/span><\/span> c&#8217;est-\u00e0-dire qu&#8217;\u00e9tant donn\u00e9 une collection finie d&#8217;expressions <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1, F_2, \\cdots, F_k<\/span><\/span> il est v\u00e9rifi\u00e9 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\">Alors nous prouverons en cons\u00e9quence qu&#8217;il est valable<\/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\">En utilisant la d\u00e9finition de la conjonction, nous avons 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;\">Sur cette expression, nous pouvons appliquer les lois de De Morgan (la loi usuelle sur deux termes) pour obtenir :<\/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;\">Maintenant, si nous appliquons l&#8217;hypoth\u00e8se d&#8217;induction, nous obtiendrons :<\/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;\">Et pour cette raison, l&#8217;induction est compl\u00e8te et la propri\u00e9t\u00e9 est valable pour tout <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> en g\u00e9n\u00e9ral. La deuxi\u00e8me relation peut \u00eatre obtenue de mani\u00e8re compl\u00e8tement analogue, je la laisse donc comme exercice au lecteur, haha!<\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h2>Forme G\u00e9n\u00e9ralis\u00e9e des Lois Distributives<\/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 mani\u00e8re similaire aux lois de De Morgan<\/strong><\/a>, les lois de distribution peuvent \u00eatre g\u00e9n\u00e9ralis\u00e9es de la mani\u00e8re suivante. Soient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{F_1, \\cdots, F_n\\}<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{G_1,\\cdots, G_m\\}<\/span><\/span> deux ensembles finis d&#8217;expressions quelconques, alors les \u00e9quivalences suivantes sont valables :<\/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) \\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>D\u00c9MONSTRATION :<\/strong> Pour construire cette d\u00e9monstration, nous devons faire une double induction sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m.<\/span><\/span> Ensuite, je ferai d&#8217;abord l&#8217;induction sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> puis sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m<\/span><\/span> pour l&#8217;expression <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;\">Si nous prenons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=1,<\/span><\/span> alors cette expression se r\u00e9duit \u00e0 :<\/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) \\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;\">Ce qui est \u00e9quivalent \u00e0 dire :<\/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 \\right] \\equiv \\left[\\bigwedge_{i=1}^n\\left( F_i\\vee G_1 \\right) \\right].<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Nous allons maintenant prouver cette expression par induction sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Si nous prenons <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1,<\/span><\/span> alors l&#8217;expression se r\u00e9duit \u00e0 :<\/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;\">Ce qui, nous le savons d\u00e9j\u00e0, est vrai. Supposons maintenant qu&#8217;elle soit vraie pour un <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k<\/span><\/span> quelconque; c&#8217;est-\u00e0-dire que l&#8217;hypoth\u00e8se d&#8217;induction sera :<\/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 \\right] \\equiv \\left[\\bigwedge_{i=1}^k\\left( F_i\\vee G_1 \\right) \\right].<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Ensuite, nous montrerons en cons\u00e9quence qu&#8217;elle est vraie pour un <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k+1.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Par la d\u00e9finition de la conjonction, on a :<\/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 \\right] := \\left[\\left(\\left(\\bigwedge_{i=1}^{k}F_i \\right)\\wedge F_{k+1} \\right) \\vee G_1 \\right] <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Maintenant, en utilisant la <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vee<\/span><\/span>-distribution, on aura :<\/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 \\right] \\equiv \\left[\\left(\\left(\\bigwedge_{i=1}^{k}F_i \\right)\\vee G_{1} \\right) \\wedge \\left(F_{k+1} \\vee G_1 \\right) \\right] <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Et juste \u00e0 ce stade, nous pouvons utiliser l&#8217;hypoth\u00e8se d&#8217;induction pour obtenir :<\/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 \\right] \\equiv \\left[\\left(\\bigwedge_{i=1}^k\\left( F_i\\vee G_1 \\right) \\right) \\wedge \\left(F_{k+1} \\vee G_1 \\right) \\right] := \\left[\\bigwedge_{i=1}^{k+1}(F_{i}\\vee G_1 \\right] <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Par cons\u00e9quent, nous avons prouv\u00e9 par induction que pour tout <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{N}<\/span><\/span> il est vrai 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 G_1\\right] \\equiv \\left[\\bigwedge_{i=1}^n(F_i\\vee G_1)\\right]<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">En compl\u00e9tant l&#8217;induction sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span>, nous avons confirm\u00e9 que cela fonctionne pour le cas initial o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=1,<\/span><\/span> maintenant il ne reste plus qu&#8217;\u00e0 compl\u00e9ter l&#8217;induction sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m.<\/span><\/span> Pour ce faire, \u00e9tablissons l&#8217;hypoth\u00e8se d&#8217;induction pour un <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=l<\/span><\/span>, c&#8217;est-\u00e0-dire qu&#8217;elle fonctionne pour :<\/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 \\right) \\right] \\equiv \\left[\\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^l(F_i\\vee G_j) \\right) \\right]<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">et \u00e0 partir de cela, nous prouverons qu&#8217;elle fonctionne aussi pour <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=l+1.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">En partant, comme toujours, de la d\u00e9finition de la conjonction, on a :<\/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 \\right) \\right] := \\left[ \\left(\\bigwedge_{i=1}^n F_i \\right) \\vee \\left(\\left(\\bigwedge_{j=1}^{l} G_j \\right) \\wedge G_{l+1}\\right) \\right] <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Maintenant, en utilisant la <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vee<\/span><\/span>-distribution, on aura :<\/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 \\right) \\right] \\equiv \\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} \\right) \\right] <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Par cons\u00e9quent, en utilisant l&#8217;hypoth\u00e8se d&#8217;induction, vous pourrez \u00e9crire :<\/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 \\right) \\right] \\equiv \\left[ \\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^l(F_i\\vee G_j) \\right) \\wedge \\left( \\left( \\bigwedge_{i=1}^n F_i \\right)\\vee G_{l+1} \\right) \\right] <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Et maintenant, en prenant le r\u00e9sultat de l&#8217;induction sur <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 \\right) \\right] \\equiv \\left[ \\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^l(F_i\\vee G_j) \\right) \\wedge \\left( \\bigwedge_{i=1}^n (F_i \\vee G_{l+1} )\\right) \\right] <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Ce qui, finalement, par la d\u00e9finition de la conjonction, nous donne :<\/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 \\right) \\right] \\equiv \\left[ \\bigwedge_{i=1}^n\\left(\\bigwedge_{j=1}^{l+1}(F_i\\vee G_j) \\right) \\right] <\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Et par cons\u00e9quent, l&#8217;induction sur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m<\/span><\/span> est compl\u00e8te et l&#8217;expression :<\/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;\">est valable pour tout <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;\">Ce parcours \u00e0 travers les preuves par induction a montr\u00e9 comment des techniques rigoureuses de d\u00e9monstration math\u00e9matique peuvent \u00eatre appliqu\u00e9es non seulement dans le domaine des nombres naturels, mais aussi en logique propositionnelle. Gr\u00e2ce \u00e0 l&#8217;induction, nous avons pu \u00e9tablir la validit\u00e9 des formes g\u00e9n\u00e9ralis\u00e9es des lois de De Morgan et des lois distributives, renfor\u00e7ant ainsi la compr\u00e9hension des fondements logiques qui sous-tendent diverses disciplines math\u00e9matiques. Cette approche est essentielle non seulement pour d\u00e9velopper des comp\u00e9tences en raisonnement abstrait, mais aussi comme un outil puissant pour aborder des probl\u00e8mes complexes en math\u00e9matiques et au-del\u00e0.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Preuves par induction : R\u00e8gles g\u00e9n\u00e9ralis\u00e9es de De Morgan et Distribution R\u00c9SUM\u00c9Dans ce cours, nous abordons le sujet des preuves par induction en math\u00e9matiques et en logique propositionnelle. Nous expliquons les deux types de preuves existants : les preuves internes ou d\u00e9ductives, qui reposent sur les r\u00e8gles de la logique, et les preuves externes ou [&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":16,"footnotes":""},"categories":[617,631,569],"tags":[],"class_list":["post-28272","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-logique-mathematique","category-logique-propositionnelle","category-mathematiques"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Preuves par induction : G\u00e9n\u00e9ralisation de De Morgan et Distribution - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Tout sur comment r\u00e9aliser des preuves par induction en math\u00e9matiques et en logique. 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