(b) $$\begin{array}[t]{lcl@{\quad(\hskip1.5in)}} (p\wedge q)\Rightarrow r &\equiv& \overline{p\wedge q}\vee r \\ &\equiv& (\overline{p}\vee\overline{q})\vee r \\ &\equiv& \overline{p}\vee(\overline{q}\vee r) \\ &\equiv& p\Rightarrow(\overline{q}\vee r) \end{array}$$, (c) $$\begin{array}[t]{lcl@{\quad(\hskip1.5in)}} (p\Rightarrow\overline{q}) \wedge (p\Rightarrow\overline{r}) &\equiv& (\overline{p}\vee\overline{q}) \wedge (\overline{p}\vee\overline{r}) \\ &\equiv& \overline{p}\vee(\overline{q}\wedge\overline{r}) \\ &\equiv& \overline{p}\vee\overline{q\vee r} \\ &\equiv& \overline{p\wedge(q\vee r)} \end{array}$$, Exercise $$\PageIndex{6}\label{ex:logiceq-06}$$. $$p \Rightarrow q \equiv \overline{p} \vee q$$. - Use the truth tables method to determine whether p! We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. If quadrilateral $$ABCD$$ is both a rectangle and a rhombus. \end{array}\), Distributive laws: $$\begin{array}[t]{l} p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r), \\ p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r). Ask Question Asked 3 years, 7 months ago. The following tables summarize those rules. Subtraction is not commutative, because it is not always true that \(x-y=y-x$$. It is easy to verify with a truth table. Example $$\PageIndex{2}\label{eg:logiceq-02}$$. Given three numbers 5, 7, and 4, in that order, how should we carry out two subtractions? So the double implication is trueif P and Qare both trueor if P and Qare both false; otherwise, the double implication is false. For starters, let's look at the truth table for 'A', '-A', … If quadrilateral $$ABCD$$ is not a rectangle or not a rhombus. $$p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$$. $\begin{array}[t]{|c|c|c|c|c|c|} \hline p & q & p\Rightarrow q & \overline{q} & \overline{p} & \overline{q}\Rightarrow\overline{p} \\ \hline \text{T} & \text{T} &&&& \\ \text{T} & \text{F} &&&& \\ \text{F} & \text{T} &&&& \\ \text{F} & \text{F} &&&& \\ \hline \end{array}$, hands-on exercise $$\PageIndex{3}\label{he:logiceq-03}$$, The logical connective exclusive or, denoted $$p\veebar q$$, means either $$p$$ or $$q$$ but not both. Laws of the excluded middle, or inverse laws: Any statement is either true or false, hence $$p\vee\overline{p}$$ is always true. But the logical equivalences $$p\vee p\equiv p$$ and $$p\wedge p\equiv p$$ are true for all $$p$$. Two propositions p and q arelogically equivalentif their truth tables are the same. The compound statement $$p\wedge\overline{p}$$ claims that $$p$$ is true, and at the same time, $$\overline{p}$$ is also true (which means $$p$$ is false). This truth table describes precisely when p^q is true (or false). Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Which ones are tautologies? Since $$\wedge$$ and $$\vee$$ are binary operations, we can only work on a pair of statements at a time. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Commutative properties apply to operations on two logical statements, but associative properties involves three logical statements. �_���}�������5~x�T>�ׯ�O�ݿ��W�|Ω������}K�����!|n����+�Z[��Ϗ8|N�1_�{�9�־����9_-�{#������_�ϡ�����j|��s�x�Ϯ~������hx����s�x�+RDÿ��b"�? p q p^q T T T T F F F T F F F F Figure 1.3 A truth table for conjunction. verify logical equivalence without truth table. If quadrilateral $$ABCD$$ is not a rectangle and it is not a rhombus. x��[�%9������W��*A���I� X�S�1�E��k8�5�����;����3��_�sH/7��������������?����$������g_J�����_���x����J of 45 degrees, then $$ABC$$ is not a right triangle. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. Use truth tables to verify the following equivalent statements. P Q P ↔ Q T T T T F F F T F F F T You should remember — or be able to construct — the truth tables for the logical connectives. hands-on exercise $$\PageIndex{2}\label{he:logiceq-02}$$. Since $$p$$ and $$q$$ represent two different statements, they cannot be the same. We illustrate how to use De Morgan’s … Compare this to the equation $$x^2=x$$, where $$x$$ is a real number. p q p^q T T T T F F F T F F F F Figure 1.3 A truth table for conjunction. Have questions or comments? 5 0 obj Example $$\PageIndex{4}\label{eg:logiceq-04}$$. The answer is: it does not matter. Commutative properties: $$\begin{array}[t]{l} p \vee q \equiv q \vee p, \\ p \wedge q \equiv q \wedge p. \end{array}$$, Associative properties: $$\begin{array}[t]{l} (p \vee q) \vee r \equiv p \vee (q \vee r), \\ (p \wedge q) \wedge r \equiv p \wedge (q \wedge r). hands-on exercise \(\PageIndex{4}\label{he:logiceq-04}$$, Since $$0\leq x\leq 1$$ means “$$x\geq 0$$ and $$x\leq 1$$,” its negation should be “$$x<0$$ or $$x>1$$”. ��wB/�s}������{�c/).�L��Ќ�k͂s%4��e*\���(�7�����ZZ�iK�N)���ii��4T��G�u�ǚgۈCDM����ǩ���V�ݩ�(�(h0���fj����j1����S����������~&�Ͼ�8�3k�L�����ן��'���O� �+ԫ�BL�����>��m���{�ܠ��t�۠Jϕ ʂ���s[ק�N���Uۗ�Օm�F�\�0��t��qY�QJRxs6ț�uk\���J>͡���U ���p���vk����.�n�O��-����A����)�޶��8ۮ�M���v�A����)�Ԛk*��]˹ژ�>��7mc��iA>���T/���q�A��e���� ?��SW|v��=�TZ4���l�긪�j�Ȅҩ1�����z�C%꫟�ܤ���?�^�P�4U8f�_� ���� }]��9.�P�qr�������ꤘr9�������A���H�v�;�v���뛖&_�s�Ҹ��? - Use the truth tables method to determine whether the formula ’: p^:q!p^q is a logical consequence of the formula : :p. Solution. Likewise, a statement cannot be both true and false at the same time, hence $$p\wedge\overline{p}$$ is always false. <> The truth tables for (a) and (b) are depicted below. Symbolically, the argument says $[(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})]. If quadrilateral $$ABCD$$ is a square, then it is both a rectangle and a rhombus. Show that $$(p \Rightarrow q) \Leftrightarrow (\overline{q} \Rightarrow \overline{p})$$ is a tautology. A tautology is a proposition that is always true, regardless of the truth values of the propositional variables it contains. Hence, $$p\wedge \overline{p}$$ must be false. Commutative properties: In short, they say that “the order of operation does not matter.” It does not matter which of the two logical statements comes first, the result from conjunction and disjunction always produces the same truth value. Construct a truth table for each formula below. We have set up the table for (a), and leave the rest to you. In the first column for the truth values of $$p$$, fill the upper half with T and the lower half with F. In the next column for the truth values of $$q$$, repeat the same pattern, separately, with the upper half and the lower half. We have the following properties for any propositional variables $$p$$, $$q$$, and $$r$$. Identity laws: Compare them to the equation $$x\cdot1=x$$: the value of $$x$$ is unchanged after multiplying by 1. Use a truth table to verify the De Morgan’s law $$\overline{p\vee q} \equiv \overline{p}\wedge\overline{q}$$. Use only the properties of logical equivalences to verify (b) and (c) in Problem 4. Exercises Give a logical explanation as well as a graphical explanation. \end{array}\), Idempotent laws: $$\begin{array}[t]{l} p \vee p \equiv p, \\ p \wedge p \equiv p. \end{array}$$, De Morgan’s laws: $$\begin{array}[t]{l} \overline{p\vee q} \equiv \overline{p}\wedge\overline{q}, \\ \overline{p\wedge q} \equiv \overline{p}\vee \overline{q}. So we split the upper half of the second column into two halves, fill the top half with T and the lower half with F. Likewise, split the lower half of the second column into two halves, fill the top half with T and the lower half with F. Repeat the same pattern with the third column for the truth values of \(r$$, and so on if we have more propositional variables. Proving logical equivalence without truth tables. This is followed by the “outer” operation to complete the compound statement. Exercise $$\PageIndex{3}\label{ex:logiceq-03}$$. Idempotent laws: When an operation is applied to a pair of identical logical statements, the result is the same logical statement. Use a truth table to show that \[[(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})]$ is a tautology. T stands for a tautology & F stands for a contradiction. Example $$\PageIndex{3}\label{eg:logiceq-03}$$, “If $$p$$ and $$q$$, then $$r$$. Simplify to an equivalent expression that is a single letter (T, F, p or ~p ). p q :p p^:q p^q p^:q!p^q T T F F T T T F F T F F F T T F F T F F T F F T j= ’since each interpretation satisfying psisatisﬁes also ’.] Consequently, $$p\equiv q$$ is same as saying $$p\Leftrightarrow q$$ is a tautology. Logical Equivalence ! Example $$\PageIndex{8}\label{eg:logiceq-10}$$. Properties of Logical Equivalence. They change from inclusion to exclusion when we take negation. $$\overline{p\Leftrightarrow q} \equiv \overline{p} \Leftrightarrow \overline{q}$$, $$(p\Rightarrow q) \vee (p\Rightarrow\overline{q}) \equiv \overline{p}$$, $$(p\Rightarrow q)\Rightarrow r \equiv p\Rightarrow (q\Rightarrow r)$$, $$p\Rightarrow (q\vee r) \equiv (p\Rightarrow q) \vee (p\Rightarrow r)$$, $$p\Rightarrow (q\wedge r) \equiv (p\Rightarrow q) \wedge (p\Rightarrow r)$$, $$(p\Rightarrow\overline{q}) \wedge (p\wedge q)$$, $$(p\Rightarrow\overline{q}) \wedge (\overline{q}\Rightarrow p)$$. hands-on exercise $$\PageIndex{1}\label{he:logiceq-01}$$. Logical Rules of Reasoning: At the foundation of formal reasoning and proving lie basic rules of logical equivalence and logical implications. Active 3 years, 1 month ago. The truth table for the conjunc-tion of two statements is shown in Figure 1.3. Equivalence of an implication and its contrapositive: $$p \Rightarrow q \equiv \overline{q} \Rightarrow \overline{p}$$. Let us explain them in words, and compare them to similar operations on the real numbers. It may not be easy to memorize the names of all these properties; however, they should all make sense to you. Determine whether the following formulas are tautologies, contradictions, or neither: Exercise $$\PageIndex{13}\label{ex:logiceq-13}$$, (a) $$p\wedge q$$ It is important to remember that $\overline{p\Rightarrow q} \not\equiv q\Rightarrow p,$ and $\overline{p\Rightarrow q} \not\equiv \overline{p}\Rightarrow\overline{q}$ either. 2. 0. equivalence. $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, [ "article:topic", "authorname:hkwong", "license:ccbyncsa", "showtoc:yes" ], $$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, $\begin{array}{|c|c|c|c|} \hline p & \overline{p} & p \vee \overline{p} & p \wedge \overline{p} \\ \hline \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{F} \\ \hline \end{array}$, $\begin{array}{|*{7}{c|}} \hline p & q & p\Rightarrow q & \overline{q} & \overline{p} & \overline{q}\Rightarrow\overline{p} & (p \Rightarrow q) \Leftrightarrow (\overline{q} \Rightarrow \overline{p}) \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} & \text{F} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \hline \end{array}$, \[[(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})]. 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