�VK����)��SXCu��S��}�8�u���a���3� 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). The logical equivalence of statement forms P and Q is denoted by writing P Q. We have used a truth table to verify that \[[(p \wedge q) \Rightarrow r] \Rightarrow [\overline{r} \Rightarrow (\overline{p} \vee \overline{q})]\] is a tautology. The truth table for the conjunc-tion of two statements is shown in Figure 1.3. It is true only when \(x=0\) or \(x=1\). Beside distributive and De Morgan’s laws, remember these two equivalences as well; they are very helpful when dealing with implications. P ↔ Q means that P and Qare equivalent. 1. Chapter 01: Mathematical Logic Subtopics 1.1 Statement 1.2 Logical Connectives, Compound Statements and Truth Tables 1.3 Statement Pattern and Logical Equivalence Tautology, Contradiction and Contingency 1.4 Quantifiers and Quantified Statements 1.5 … Construct a truth table for each formula below. Propositional Logic Exercise 2.6. The important consequence of the associative property is: since it does not matter on which pair of statements we should carry out the operation first, we can eliminate the parentheses and write, for example, \[p\vee q\vee r\] without worrying about any confusion. Hence, \(p\wedge \overline{p}\) must be false. If \(n>1\) is prime, then \(n+1\) is composite. Not all operations are associative. LOGICAL EQUIVALENCE I introduced logic as the science of arguments. If triangle \(ABC\) is isosceles and contains an angle of 45 degrees, then \(ABC\) is a right triangle. Consequently, \(p\equiv q\) is same as saying \(p\Leftrightarrow q\) is a tautology. Two logical statements are logically equivalent if they always produce the same truth value. \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). Proving logical equivalence without truth tables. Thus, ([eqn:tautology]) cannot be false, it must be a tautology. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to \(T\). (b) \( \begin{array}[t]{[email protected]{\quad(\hskip1.5in)}} (p\wedge q)\Rightarrow r &\equiv& \overline{p\wedge q}\vee r \\ &\equiv& (\overline{p}\vee\overline{q})\vee r \\ &\equiv& \overline{p}\vee(\overline{q}\vee r) \\ &\equiv& p\Rightarrow(\overline{q}\vee r) \end{array}\), (c) \( \begin{array}[t]{[email protected]{\quad(\hskip1.5in)}} (p\Rightarrow\overline{q}) \wedge (p\Rightarrow\overline{r}) &\equiv& (\overline{p}\vee\overline{q}) \wedge (\overline{p}\vee\overline{r}) \\ &\equiv& \overline{p}\vee(\overline{q}\wedge\overline{r}) \\ &\equiv& \overline{p}\vee\overline{q\vee r} \\ &\equiv& \overline{p\wedge(q\vee r)} \end{array}\), Exercise \(\PageIndex{6}\label{ex:logiceq-06}\). If triangle \(ABC\) is not a right triangle, then \(ABC\) is not isosceles. This requires \(p\) to be true and \(q\) to be false, which translates into \(p \wedge \overline{q}\). �_���}�������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"�? Instead, since \(p\Rightarrow q \equiv \overline{p}\vee q\), it follows from De Morgan’s law that \[\overline{p \Rightarrow q} \equiv \overline{\overline{p} \vee q} \equiv p \wedge \overline{q}.\] Alternatively, we can argue as follows. 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. We have the following properties for any propositional variables \(p\), \(q\), and \(r\). <> Exercise \(\PageIndex{1}\label{ex:logiceq-01}\). We call the number 1 the multiplicative identity. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to \(T\). Example \(\PageIndex{3}\label{eg:logiceq-03}\), “If \(p\) and \(q\), then \(r\). Which of the following statements are equivalent to the statement “if \(x^2>0\), then \(x>0\)”? It says that \(p \Rightarrow q\) is true when one of these two things happen: (i) when \(p\) is false, (ii) otherwise (when \(p\) is true) \(q\) must be true. The answer is: it does not matter. 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 ’.] ... and can often be used to prove logical equivalences without the use of truth tables. We have learned that \[p\Leftrightarrow q \equiv (p\Rightarrow q) \wedge (q\Rightarrow p),\] which is the reason why we call \(p\Leftrightarrow q\) a biconditional 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. Two propositions p and q arelogically equivalentif their truth tables are the same. ����sMz�a�~���a��襖C��sS4�Y�c�6ߖ5���"'�s��T���R1��K��о^$8��� �ޥ�P��_P`�t6\gq�S �C[$���"��T$�\�C��2��_�H,��������[����m-�o�Ty�V�7 �D� �-tȹ�t�H�Y.�8�"��~4cx��c4��U��7Y�G�}����|%�0����}~,ts߱�o���>�/?�}?ĝ�b�,�Θ��wSe�}v!���� ����=��eծh_yR;��oơ~ԅ~�8�0�0c�7�Y��o�죣���9ܴ~.��śn�m6����D�Q�4P1Z��~�)v \(p \Rightarrow q \equiv \overline{p} \vee q\). For starters, let's look at the truth table for 'A', '-A', … The truth table for the conjunc-tion of two statements is shown in Figure 1.3. In other words, show that the logic used in the argument is correct. then \(ABCD\) has two sides of equal length. - Use the truth tables method to determine whether p! We need eight combinations of truth values in \(p\), \(q\), and \(r\). If quadrilateral \(ABCD\) is a square, then it is both a rectangle and a rhombus. Distributive laws: When we mix two different operations on three logical statements, one of them has to work on a pair of statements first, forming an “inner” operation. Use truth tables to verify these logical equivalences. 3 5 Truth Table Method We can check for logical entailment by comparing tables of all possible interpretations. (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.) Be sure to fill them in. We can also argue that this compound statement is always true by showing that it can never be false. If quadrilateral \(ABCD\) is not a rectangle and it is not a rhombus. (c) \(p\wedge q\), Exercise \(\PageIndex{14}\label{ex:logiceq-14}\). Determine whether the following formulas are tautologies, contradictions, or neither: Exercise \(\PageIndex{13}\label{ex:logiceq-13}\), (a) \(p\wedge q\) The negation of an implication: \(\overline{p \Rightarrow q} \equiv p \wedge \overline{q} \). Example \(\PageIndex{4}\label{eg:logiceq-04}\). \((p\Rightarrow q) \vee (p\Rightarrow \overline{q})\), \((p\wedge q)\Leftrightarrow p \equiv p\Rightarrow q\), \((p\wedge q)\Rightarrow r \equiv p\Rightarrow(\overline{q}\vee r)\), \((p\Rightarrow\overline{q}) \wedge (p\Rightarrow\overline{r}) \equiv \overline{p\wedge(q\vee r)}\). Commutative properties apply to operations on two logical statements, but associative properties involves three logical statements. Note that all of those rules can be proved using truth tables. Section 1.1: Logical Form and Logical Equivalence An argument is a sequence of statements aimed at demonstrating the truth of an assertion. Active 2 years, 2 months ago. This truth table describes precisely when p^q is true (or false). Distributive laws say that we can distribute the “outer” operation over the inner one. They change from inclusion to exclusion when we take negation. The following tables summarize those rules. Take note of the two endpoints 2 and 3. We have set up the table for (a), and leave the rest to you. Associative properties: Roughly speaking, these properties also say that “the order of operation does not matter.” However, there is a key difference between them and the commutative properties.

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