{"id":28160,"date":"2021-04-09T13:00:36","date_gmt":"2021-04-09T13:00:36","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28160"},"modified":"2024-09-03T17:49:23","modified_gmt":"2024-09-03T17:49:23","slug":"induction-proofs-generalized-de-morgan-and-distribution-rules","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/induction-proofs-generalized-de-morgan-and-distribution-rules\/","title":{"rendered":"Induction Proofs: Generalized De Morgan and Distribution Rules"},"content":{"rendered":"<p><center><\/p>\n<h1>Induction Proofs: Generalized Rules of De Morgan and Distribution<\/h1>\n<p><\/p>\n<p style=\"text-align:center;\"><strong>SUMMARY<\/strong><br \/><em>This class addresses the topic of induction proofs in mathematics and propositional logic. The two types of existing proofs are explained: internal or deductive proofs, which are based on the rules of logic, and external or metamathematical proofs, which are necessary to prove statements that refer to logic itself. Mathematical Induction is introduced as a demonstration method that allows proving that certain statements hold for all natural numbers. An example is presented with the corresponding proof, and the generalized forms of De Morgan&#8217;s laws and the distributive laws in propositional logic are explained, along with their respective induction proofs. This class is of great importance for understanding the fundamentals of mathematics and logic and for applying them in different areas of knowledge.<\/em><\/p>\n<p><\/center><br \/>\n<\/p>\n<p style=\"text-align:center;\"><strong>LEARNING OBJECTIVES:<\/strong><br \/>\nBy the end of this class, the student will be able to:\n<\/p>\n<ol>\n<li><strong>Recognize<\/strong> the two types of proofs to be distinguished: internal or deductive proofs and external or metamathematical proofs.<\/li>\n<li><strong>Apply<\/strong> mathematical induction to perform proofs involving natural numbers and propositional logic.<\/li>\n<li><strong>Use<\/strong> conjunction and disjunction notations to write expressions in propositional logic.<\/li>\n<li><strong>Understand<\/strong> the generalized form of De Morgan&#8217;s laws and the Distributive Laws in Propositional Logic.<\/li>\n<li><strong>Understand<\/strong> the concept of the induction hypothesis and its role in the induction proof.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong>INDEX<\/strong><br \/>\n<a href=\"#1\">INTERNAL AND EXTERNAL PROOFS<\/a><br \/>\n<a href=\"#2\">MATHEMATICAL INDUCTION PROOFS<\/a><br \/>\n<a href=\"#3\">INDUCTION PROOFS IN PROPOSITIONAL LOGIC<\/a><br \/>\n<a href=\"#4\">GENERALIZED FORM OF DE MORGAN&#8217;S LAWS<\/a><br \/>\n<a href=\"#5\">GENERALIZED FORM OF THE DISTRIBUTIVE LAWS<\/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>Internal and External Proofs<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=212s\" target=\"_blank\" rel=\"noopener\"><strong>There are two types of proofs that must be distinguished.<\/strong><\/a> So far, we have seen many examples of formal proofs. These types of proofs emerge from the rules of logic. Such proofs are said to take place \u00abwithin logic,\u00bb and therefore they are also referred to as \u00abinternal\u00bb or deductive proofs. These formal proofs have a limited scope because they only serve to prove statements that can be written in the language of logic. However, we might want to prove some things about logic itself. We might want to prove that all statements of propositional logic satisfy a certain property. Such statements, which refer to logic itself, cannot be established or proved within logic. To prove such statements, we use an external proof. External proofs are sometimes called \u00abmetamathematical,\u00bb and we have already encountered such things, like when we saw the (meta)deduction theorem. It is here that we contextualize inductive proofs.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Mathematical Induction Proofs<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=359s\" target=\"_blank\" rel=\"noopener\"><strong>Mathematical Induction is a method of proof<\/strong><\/a> that allows us to prove that some things hold for all natural numbers.<\/p>\n<p style=\"text-align: justify;\"><strong>EXAMPLE:<\/strong> It is possible to prove that any number of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^n - 4^n<\/span><\/span>, where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> is any natural number, is always divisible by 7.<br \/>\n<strong>PROOF:<\/strong> If we observe what happens with <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1<\/span><\/span>, we will see that:<\/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;\">which is obviously divisible by 7.<\/p>\n<p style=\"text-align: justify;\">Now suppose that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^n - 4^n<\/span><\/span> is divisible for a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k.<\/span><\/span> From this, we will prove that this expression will also hold for <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k+1.<\/span><\/span> We can do this as follows:<\/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;\">Therefore, if <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^k - 4^k<\/span><\/span> is divisible by 7, then <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11 ( 11^{k} - 4^{k} ) + 7\\cdot 4^{k}<\/span><\/span> will also be divisible by 7, which is the same as saying that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^{k+1} - 4^{k+1}<\/span><\/span> is divisible by 7. From here, we have that if <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">11^k - 4^k<\/span><\/span> is divisible for <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=1<\/span><\/span>, then it will be for <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=2, k=3, k=4,\\cdots<\/span><\/span> and so on, and therefore divisible for any <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{N}.<\/span><\/span> When this occurs, we say that the induction is complete. \u25a0<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Induction Proofs in Propositional Logic<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=775s\" target=\"_blank\" rel=\"noopener\"><strong>For the induction proofs that we will perform next,<\/strong><\/a> it will first be necessary to introduce the following notation convention<\/p>\n<p style=\"text-align: justify;\"><strong>NOTATION:<\/strong> Let <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1,\\cdots, F_n<\/span><\/span> be any finite set of expressions in propositional logic. The conjunctions and disjunctions of these expressions are introduced as follows:<\/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;\">With this, we can now deal with the following two generalized forms.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Generalized Form of De Morgan&#8217;s Laws<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=829s\" target=\"_blank\" rel=\"noopener\"><strong>Given a finite set of expressions in propositional logic<\/strong><\/a> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1,\\cdots, F_n,<\/span><\/span> the following two properties will always hold:<\/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>PROOF:<\/strong> First, we will prove by induction on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> that <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;\">First, we must check what happens with the initial case <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1.<\/span><\/span> In this case, it is clear that <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;\">Now suppose the property works for some <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k;<\/span><\/span> that is, given a finite collection of expressions <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1, F_2, \\cdots, F_k<\/span><\/span>, it holds that:<\/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\">Then we will prove that it consequently holds<\/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\">Using the definition of conjunction, we have that:<\/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;\">On this expression, we can apply De Morgan&#8217;s laws (the usual one on two terms) to obtain:<\/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;\">Now, if we apply the induction hypothesis, we obtain:<\/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;\">And for this reason, the induction is complete, and the property holds for any <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> in general. The second relation can be obtained in a completely analogous way, so I&#8217;ll leave it as an exercise for the reader, mwahaha!<\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h2>Generalized Form of the Distributive Laws<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=eJQcNPrKyW0&amp;t=1205s\" target=\"_blank\" rel=\"noopener\"><strong>Similar to De Morgan&#8217;s laws<\/strong><\/a>, the distributive laws can be generalized as follows. Let <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{F_1, \\cdots, F_n\\}<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{G_1,\\cdots, G_m\\}<\/span><\/span> be two finite sets of any expressions, then the following equivalences hold:<\/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>PROOF:<\/strong> To build this proof, we must perform a double induction on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m.<\/span><\/span> Next, I will do the induction first on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span> and then on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m<\/span><\/span> for the 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;\">If we take <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=1,<\/span><\/span> then this expression is written as<\/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;\">Which is the same as saying:<\/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;\">Now we will prove this expression by induction on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">If we take <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=1,<\/span><\/span> then the expression reduces to<\/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;\">Which we already know is true. Now suppose it holds for some <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k<\/span><\/span>; that is, the induction hypothesis will be:<\/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;\">Then, from this, we will show that consequently, it holds for <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n=k+1.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">By the definition of conjunction, we have that:<\/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;\">Now, using the <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vee<\/span><\/span>-distribution, we have:<\/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;\">And right at this point, we can use the induction hypothesis to obtain:<\/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;\">Therefore, we have proved by induction that for all <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{N}<\/span><\/span>, it holds that<\/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;\">Completing the induction on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/span>, we have verified that the initial case for <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=1,<\/span><\/span> works. Now we just need to complete the induction on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m.<\/span><\/span> To do the latter, we establish the induction hypothesis for a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=l<\/span><\/span>, that is, it works:<\/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;\">and from this, we will prove that it also works for <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=l+1.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Starting, as always, from the definition of conjunction, we have that<\/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;\">Now, using the <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vee<\/span><\/span>-distribution, we have<\/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;\">Consequently, using the induction hypothesis, you can write<\/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;\">And if we now take the result of the induction on <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;\">Which, finally, by the definition of conjunction, we have<\/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;\">And therefore, the induction on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m<\/span><\/span> is complete, and the 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;\">is valid for all <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;\">This exploration of induction proofs has demonstrated how rigorous mathematical demonstration techniques can be applied not only in the realm of natural numbers but also in propositional logic. Through induction, we have established the validity of the generalized forms of De Morgan&#8217;s laws and the distributive laws, thereby reinforcing the understanding of the logical foundations underlying various areas of mathematical knowledge. This approach is not only essential for the development of abstract reasoning skills but also serves as a powerful tool for tackling complex problems in mathematics and beyond.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Induction Proofs: Generalized Rules of De Morgan and Distribution SUMMARYThis class addresses the topic of induction proofs in mathematics and propositional logic. The two types of existing proofs are explained: internal or deductive proofs, which are based on the rules of logic, and external or metamathematical proofs, which are necessary to prove statements that refer [&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":[605,567,619],"tags":[],"class_list":["post-28160","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematical-logic","category-mathematics","category-propositional-logic"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Induction Proofs: Generalized De Morgan and Distribution Rules - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Everything about Performing Induction Proofs in Mathematics and Logic. 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