After studying tautologies, contradictions, and valid arguments, truth tables can now be used as practical tools. Many situations depend on several conditions being true or false: program loops, access rules, decision systems, and logic puzzles. Truth-table applications translate the situation into primitive statements and then evaluate the compound condition. In this lesson, truth-table applications will show how formal logic checks real decision patterns. :::definition[Truth-Table Application] A truth-table application is a use of a truth table to analyze a real situation that depends on logical conditions. Such situations include programming decisions, rule-based systems, and logic puzzles. ::: To apply a truth table, first identify the basic conditions. Next, assign a primitive statement to each condition. Then write the compound statement that represents the full rule. Finally, evaluate the statement under each possible assignment or under the assignments given in the problem. This process keeps the reasoning disciplined and prevents guessing. :::solved-problem A loop terminates when $((a>0)\land(b<5))\lor\neg((c=2)\land(d>10))$ is true. Determine whether the loop terminates for: (1) $a=3,b=4,c=2,d=12$. (2) $a=0,b=4,c=1,d=12$. (3) $a=5,b=8,c=2,d=15$. (4) $a=1,b=2,c=3,d=20$. ::: :::solution Let $$\begin{equation*} T=((a>0)\land(b<5))\lor\neg((c=2)\land(d>10)). \end{equation*}$$ [1] If $a=3,b=4,c=2,d=12$, then $(a>0)\land(b<5)$ is true and $\neg((c=2)\land(d>10))$ is false. Hence, $T=1$, so the loop terminates. [2] If $a=0,b=4,c=1,d=12$, then $(a>0)\land(b<5)$ is false and $(c=2)\land(d>10)$ is false. Its negation is true. Hence, $T=1$, so the loop terminates. [3] If $a=5,b=8,c=2,d=15$, then $(a>0)\land(b<5)$ is false and $(c=2)\land(d>10)$ is true. Its negation is false. Hence, $T=0$, so the loop does not terminate. [4] If $a=1,b=2,c=3,d=20$, then $(a>0)\land(b<5)$ is true. Hence, $T=1$, so the loop terminates. ::: Use the checker to evaluate the loop condition by smaller pieces. Enter values for $a$, $b$, $c$, and $d$, then compare the two main parts of the compound statement. The output shows whether each relational test is true or false before giving the final termination decision. This mirrors the truth-table application method: identify conditions, evaluate substatements, and then combine the truth values. :::calculator[Loop Termination Checker] ::: :::solved-problem Four students are asked who erased the board before class. Their statements are: Asha says Ben erased the board. Ben says he did not erase the board. Cira says Dev erased the board. Dev says Cira's statement is false. Given that exactly one statement is true and exactly one student erased the board, determine the student who erased the board. ::: :::solution Let $A,B,C,D$ denote the statements that Asha, Ben, Cira, and Dev erased the board respectively. The four spoken statements are $B$, $\neg B$, $D$, and $\neg D$. Since exactly one student erased the board, test each possible eraser. If Asha erased the board, then $B=0$ and $D=0$, so the spoken truth values are $0,1,0,1$. This gives two true statements. If Ben erased the board, then $B=1$ and $D=0$, so the spoken truth values are $1,0,0,1$. This gives two true statements. If Cira erased the board, then $B=0$ and $D=0$, so the spoken truth values are $0,1,0,1$. This gives two true statements. If Dev erased the board, then $B=0$ and $D=1$, so the spoken truth values are $0,1,1,0$. This gives two true statements. Therefore, no possible eraser makes exactly one statement true. Hence, the conditions as stated are inconsistent, and the puzzle has no solution. ::: :::remark The logic puzzle above illustrates an important mathematical habit: when a problem's conditions are inconsistent, the correct conclusion is not to force an answer. A truth-table application may show that no assignment satisfies all the stated conditions. This is a legitimate result, not a failure of the method. ::: :::solved-problem A security rule grants access when $(p\land q)\lor r$ is true, where $p$ means the password is correct, $q$ means the device is registered, and $r$ means an administrator override is active. Determine whether access is granted for $(p,q,r)=(1,0,0)$, $(1,1,0)$, and $(0,0,1)$. ::: :::solution Let $S=(p\land q)\lor r$. For $(p,q,r)=(1,0,0)$, $$\begin{align*} S&=(1\land 0)\lor 0\\ &=0\lor 0\\ &=0. \end{align*}$$ Access is not granted. For $(p,q,r)=(1,1,0)$, $$\begin{align*} S&=(1\land 1)\lor 0\\ &=1\lor 0\\ &=1. \end{align*}$$ Access is granted. For $(p,q,r)=(0,0,1)$, $$\begin{align*} S&=(0\land 0)\lor 1\\ &=0\lor 1\\ &=1. \end{align*}$$ Access is granted. ::: :::exercise A notification is sent when $(p\lor q)\land\neg r$ is true. Here $p$ means an assignment is due, $q$ means a quiz is scheduled, and $r$ means notifications are muted. Determine whether a notification is sent for: (1) $(p,q,r)=(1,0,0)$. (2) $(p,q,r)=(0,0,0)$. (3) $(p,q,r)=(0,1,1)$. ::: :::answer (1) $(1\lor 0)\land\neg 0=1\land 1=1$, so a notification is sent. (2) $(0\lor 0)\land\neg 0=0\land 1=0$, so no notification is sent. (3) $(0\lor 1)\land\neg 1=1\land 0=0$, so no notification is sent. ::: :::faq[Frequently Asked Questions] Q: What is the first step in a truth-table application? A: Identify the primitive conditions and assign symbols to them. Q: Can truth tables show that a puzzle has no solution? A: Yes. If no assignment satisfies all conditions, the correct conclusion is that the conditions are inconsistent. Q: Why are applications useful in discrete mathematics? A: They show how logical connectives control decisions in algorithms, rules, and structured reasoning. ::: :::call-to-action[Consolidate Truth-Table Applications] subtitle: Revise the full sequence of logical connectives, translations, truth tables, and applications before moving to the next section. button: Review Section | /discrete-mathematics/mathematical-logic/basic-logical-connectives-and-truth-tables/statements-and-primitive-statements button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/basic-logical-connectives-and-truth-tables/tautologies-contradictions-and-valid-arguments :::