Having studied logical implication and rules of inference, we now move from whole statements to statements containing variables. Open statements and universe of discourse form the entry point to predicate logic, because a sentence such as $x+2$ is an even integer cannot be judged true or false until $x$ is chosen. The same expression may behave differently when the allowed values of the variable change. In this lesson, open statements and universe of discourse will be used to assign truth values, form truth sets, and evaluate compound open statements. :::definition[Open Statement] A declarative sentence containing one or more variables is called an open statement if its truth value cannot be determined until values are assigned to its variables. If a value from the allowed set is substituted for each variable, then the open statement becomes a statement and has a definite truth value. ::: :::definition[Universe of Discourse] The universe of discourse is the set from which the variables in an open statement are allowed to take values. The truth value of an open statement depends on both the expression and the chosen universe of discourse. ::: :::definition[Notation for Open Statements] An open statement in one variable is usually denoted by $p(x)$, $q(x)$, or $r(x)$. An open statement in two variables is usually denoted by $p(x,y)$, $q(x,y)$, or $r(x,y)$. Here $x$ and $y$ are variables whose possible values are determined by the universe of discourse. ::: :::example[Open Statement in One Variable] Let the universe of discourse be the set of all integers, and let $$\begin{equation*} p(x): x+2\ \text{is an even integer}. \end{equation*}$$ Then $p(x)$ is an open statement. For $x=4$, the statement $p(4)$ says that $4+2$ is an even integer, which is true. For $x=5$, the statement $p(5)$ says that $5+2$ is an even integer, which is false. Hence, the truth value of $p(x)$ depends on the value assigned to $x$. ::: :::example[Open Statement in Two Variables] Let the universe of discourse be the set of all integers, and let $$\begin{equation*} q(x,y): y+2,\ x-y,\ x+2y\ \text{are even integers}. \end{equation*}$$ For $x=4$ and $y=2$, we get $$\begin{equation*} y+2=4,\quad x-y=2,\quad x+2y=8. \end{equation*}$$ Each number is even. Hence, $q(4,2)$ is true. ::: :::definition[Truth Set of an Open Statement] Let $p(x)$ be an open statement whose universe of discourse is $U$. The truth set of $p(x)$ is the set of all elements $a\in U$ such that $p(a)$ is true. It is written as $$\begin{equation*} \{a\in U: p(a)\ \text{is true}\}. \end{equation*}$$ ::: A truth set collects exactly the elements of the universe that make an open statement true. Choose a finite integer universe and then choose a predicate to test against every allowed value. The table evaluates the predicate one substitution at a time, and the final line gathers the successful substitutions into a set. This helps separate the open statement from its truth set, which is a set of values rather than another sentence. :::scientific-preview[Truth Set Checker] ::: :::solved-problem Let the universe of discourse be the set of all integers. Let $p(x):x\leq 3$ and $q(x):x+1$ is odd. Determine the truth values of $p(1)$, $q(1)$, $\neg p(3)$, and $q(6)$. ::: :::solution Given that $p(x):x\leq 3$ and $q(x):x+1$ is odd over the integers. For $p(1)$, we have $1\leq 3$, so $p(1)$ is true. For $q(1)$, we have $1+1=2$, and $2$ is not odd, so $q(1)$ is false. For $\neg p(3)$, first $p(3)$ is true because $3\leq 3$; hence $\neg p(3)$ is false. For $q(6)$, we have $6+1=7$, and $7$ is odd, so $q(6)$ is true. Therefore, $$\begin{equation*} p(1):T,\quad q(1):F,\quad \neg p(3):F,\quad q(6):T. \end{equation*}$$ ::: :::solved-problem Let the universe of discourse be the set of all integers. Let $p(x):x^2=2x$. Find all integers $x$ for which $p(x)$ is true. ::: :::solution Given that $p(x):x^2=2x$ over the integers. To find all integers $x$ for which $p(x)$ is true, solve $$\begin{align*} x^2&=2x\\ x^2-2x&=0\\ x(x-2)&=0. \end{align*}$$ Therefore $x=0$ or $x=2$. Hence the truth set of $p(x)$ is $\{0,2\}$. ::: :::exercise Let the universe of discourse be the set of all integers. Let $p(x):x\leq 3$, $q(x):x+1$ is odd, and $r(x):x>0$. Determine the truth values of $p(3)\lor[q(3)\lor\neg r(3)]$, $\neg p(3)\land[q(3)\lor r(3)]$, and $p(2)\to[q(2)\to r(2)]$. Find all $x$ for which $[p(x)\land q(x)]\land r(x)$ is true. ::: :::answer The first statement is true, the second is false, and the third is true. The compound statement $[p(x)\land q(x)]\land r(x)$ is true exactly for positive even integers not exceeding $3$, so its truth set is $\{2\}$. ::: :::faq[Frequently Asked Questions] Q: Why is an open statement not simply true or false? A: Its truth value depends on the value assigned to its variable. Q: Can the same open statement have different truth values in different universes? A: Yes. The universe of discourse controls which substitutions are allowed. Q: What is a truth set? A: It is the set of all elements in the universe that make the open statement true. ::: :::call-to-action[Practise Open Statements] subtitle: Evaluate open statements carefully before adding existential and universal quantifiers. button: Next Lesson | /discrete-mathematics/mathematical-logic/predicates-and-quantifiers/existential-and-universal-quantifiers button-ghost: Previous Lesson | /discrete-mathematics/mathematical-logic/logical-implication-and-rules-of-inference/invalid-arguments-and-counterexamples :::