Two logical formulas $$p$$ and $$q$$ are logically equivalent, denoted $$p\equiv q,$$ (defined in section 2.2) if and only if  $$p \Leftrightarrow q$$ is a tautology. The term contingency is not as widely used as the terms tautology and contradiction. (As before, we write the truth values for p and q in the order of TT, TF, FT, FF from top to bottom in the table.) Commutative properties apply to operations on two logical statements, but associative properties involves three logical statements. Exercises Viewed 7k times 0 $\begingroup$ $(p\land q)\rightarrow r$ and $(p\rightarrow r)\lor (q\rightarrow r)$ Have to try prove if they are logically equivalent or not using the laws listed below and also if need to use negation and implication laws. Denote by $$T$$ and $$F$$ a tautology and a contradiction, respectively. Symbolically, the argument says $[(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})]. We have set up the table for (a), and leave the rest to you. Then \[(p \wedge q) \Rightarrow r \quad \mbox{must be true}, \qquad\mbox{and}\qquad \overline{r} \Rightarrow (\overline{p} \vee \overline{q}) \quad \mbox{must be false}.$ For the second implication to be false, we need $\overline{r} \quad\mbox{to be true}, \qquad\mbox{and}\qquad \overline{p} \vee \overline{q} \quad\mbox{to be false}.$ They in turn imply that $$r$$ is false, and both $$\overline{p}$$ and $$\overline{q}$$ are false; hence both $$p$$ and $$q$$ are true. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. %PDF-1.3 <> Be sure you understand and memorize the last three equivalences, because we will use them frequently in the rest of the course. $$\begin{array}[t]{ {|c | c | c | c | c | c |}} \hline p & q & p\vee q & \overline{p\vee q} & \overline{p} & \overline{q} & \overline{p}\wedge\overline{q} \\ \hline T & T & T & F & F & F & F \\ T & F & T & F & F & T & F \\ F & T & T & F & T & F & F \\ F & F &F & T & T & T & T \\ \hline \end{array}$$, Exercise $$\PageIndex{2}\label{ex:logiceq-02}$$. Exercise $$\PageIndex{7}\label{ex:logiceq-07}$$. If quadrilateral $$ABCD$$ has two sides of equal length, then it is either a rectangle or a rhombus. then it is not a rectangle and it is not a rhombus. basic tools as I introduced them in chapter 1. Construct a truth table for each formula below. �����M�����*e̿�قP�r��灇�GƄG�u�#u�7��X����]'���������������s;磹�� h^�iV����i���_�\n^>?�Z�����?������)n0��aʡ�2��ч�?��g�t�SO~�>~wj�f����8L��\�ش�c�m|�?gD v{nrI1=vx�9�]�ǁ�O� �_�Z?�8�3� Z���gϐx�kA��Z���i���)�����?��IJ��:S���=J&�����ͱ�N����I����ٰ̀@�+�����F-N��h��Oǜ6��~8�mFjƶC��*����+=�!�����:1��׼\3�ȱ�ϓ���-�J"]�H�r����:��W�w�N���vm��o������ ��o����v��x���)AF���/F Not all operations are associative. Equivalence of an implication and its contrapositive: $$p \Rightarrow q \equiv \overline{q} \Rightarrow \overline{p}$$. Use truth tables to verify the two associative properties. Two propositions p and q arelogically equivalentif their truth tables are the same. This requires $$p$$ to be true and $$q$$ to be false, which translates into $$p \wedge \overline{q}$$. Exercises We are not saying that $$p$$ is equal to $$q$$. Use a truth table to verify the De Morgan’s law $$\overline{p\vee q} \equiv \overline{p}\wedge\overline{q}$$. Only (b) is a tautology, as indicated in the truth tables below. From the following truth table $\begin{array}{|c|c|c|c|} \hline p & \overline{p} & p \vee \overline{p} & p \wedge \overline{p} \\ \hline \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{F} \\ \hline \end{array}$ we gather that $$p\vee\overline{p}$$ is a tautology, and $$p\wedge\overline{p}$$ is a contradiction. Solving Logical equivalence & propositional logic problems without truth tables. LOGICAL EQUIVALENCE I introduced logic as the science of arguments. For logical operations, the identity for disjunction is F, and the identity for conjunction is T. Domination laws: Compare them to the equation $$x\cdot0=0$$ for real numbers: the result is always 0, regardless of the value $$x$$. This kind of proof is usually more difficult to follow, so it is a good idea to supply the explanation in each step. Logical Equivalence ! x��[�%9������W��*A���I� X�S�1�E��k8�5�����;����3��_�sH/7��������������?����\$������g_J�����_���x����J (a) $$\begin{array}[t]{|*{5}{c|}} \hline p & q & \overline{p} & \overline{p}\vee q & (\overline{p}\vee q)\Rightarrow p \\ \hline T & T & F & T & \qquad\;T \\ T & F & F & F & \qquad\;T \\ F & T & T & T & \qquad\; F \\ F & F & T & T & \qquad\; F \\ \hline \end{array}$$, (b) $$\begin{array}[t]{|*{6}{c|}} \hline p & q & p\Rightarrow q & \overline{q} & p\Rightarrow\overline{q} & (p\Rightarrow q)\vee(p\Rightarrow\overline{q}) \\ \hline T &T &T & F & F &T \\ T &F &F & T & T &T \\ F &T &T & F & T &T \\ F &F &T & T & T &T \\ \hline \end{array}$$, (c) $$\begin{array}[t]{|*{5}{c|}} \hline p & q & r & p\Rightarrow q & (p\Rightarrow q)\Rightarrow r \\ \hline T &T &T & T & \qquad\quad T \\ T & T & F & T & \qquad\quad F \\ T &F &T & F & \qquad\quad T \\ T &F &F & F & \qquad\quad T \\ F &T &T & T & \qquad\quad T \\ F &T &F & T & \qquad\quad F \\ F &F &T & T & \qquad\quad T \\ F &F &F & T & \qquad\quad F \\ \hline \end{array}$$, Exercise $$\PageIndex{4}\label{ex:logiceq-04}$$. That is why we write $$p\equiv q$$ instead of $$p=q$$. What is the negation of $$2\leq x\leq 3$$? Example $$\PageIndex{1}\label{eg:logiceq-01}$$. If $$n>1$$ is prime, then $$n+1$$ is composite. Be sure to fill them in. Without truth tables to show that an implication and it’s contrapositive are logically equivalent. Writing an implication as a disjunction: $$p \Rightarrow q \equiv \overline{p} \vee q$$. We can also argue that this compound statement is always true by showing that it can never be false. Exercise $$\PageIndex{8}\label{ex:logiceq-08}$$, Exercise $$\PageIndex{9}\label{ex:logiceq-09}$$. The negation of an implication: $$\overline{p \Rightarrow q} \equiv p \wedge \overline{q}$$. If quadrilateral $$ABCD$$ is not a square. If quadrilateral $$ABCD$$ is not a rectangle and it is not a rhombus. This claim is always true. equivalence. Show that $$(p \Rightarrow q) \Leftrightarrow (\overline{q} \Rightarrow \overline{p})$$ is a tautology. Here is a complete proof: $% \arraygap{1.2} \begin{array}{lcl@{\quad}l} [(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})] &\equiv& \overline{(p \wedge q) \Rightarrow r} \vee [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})] & \mbox{(implication as disjunction)} \\ &\equiv& \overline{(p \wedge q) \Rightarrow r} \vee [\overline{\overline{p} \vee \overline{q}} \Rightarrow r] & \mbox{(implication as disjunction)} \\ &\equiv& \overline{(p \wedge q) \Rightarrow r} \vee [(p \wedge q) \Rightarrow r] & \mbox{(De Morgan's law)} \\ &\equiv& T & \mbox{(inverse law)} \end{array}$ This is precisely what we called the left-to-right method for proving an identity (in this case, a logical equivalence). In words, $$p\vee\overline{p}$$ says that either the statement $$p$$ is true, or the statement $$\overline{p}$$ is true (that is, $$p$$ is false). Which ones are tautologies? $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, [ "article:topic", "authorname:hkwong", "license:ccbyncsa", "showtoc:yes" ], $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, $\begin{array}{|c|c|c|c|} \hline p & \overline{p} & p \vee \overline{p} & p \wedge \overline{p} \\ \hline \text{T} & \text{F} & \text{T} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{F} \\ \hline \end{array}$, $\begin{array}{|*{7}{c|}} \hline p & q & p\Rightarrow q & \overline{q} & \overline{p} & \overline{q}\Rightarrow\overline{p} & (p \Rightarrow q) \Leftrightarrow (\overline{q} \Rightarrow \overline{p}) \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} & \text{F} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{T} & \text{F} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \hline \end{array}$, \[[(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})].