Truth sets of predicates

Find truth sets for one-variable and multivariable predicates by solving within the declared universal set.

Truth Sets of One-Variable Predicates

A one-variable predicate is a sentence with a variable, defined over a universal set, that becomes true or false after a concrete value is substituted. The truth set records exactly the allowed inputs that make the predicate true.

For a predicate \(P(x)\) over universal set \(U\), write the truth set as \[T=\{x\in U\mid P(x)\}.\] The universal set is part of the definition: only elements of \(U\) are eligible for membership.

For example, let \(D(x)\) mean "\(x\) is divisible by \(2\)," and let \(U=\{1,2,\ldots,6\}\). Checking the six allowed inputs gives \(1\notin T\), \(2\in T\), \(3\notin T\), \(4\in T\), \(5\notin T\), and \(6\in T\).

Therefore the truth set is \[T=\{2,4,6\}.\]

Finding Truth Sets by Solving

To find a truth set algebraically, solve the predicate and then keep only solutions from the declared universe.

For \(P(x):x^2+2x=3\) with universal set \(\mathbb Z\), solve \[x^2+2x=3\quad\Longleftrightarrow\quad x^2+2x-3=0\quad\Longleftrightarrow\quad(x+3)(x-1)=0.\] The solutions are \(x=-3\) and \(x=1\), and both are integers, so \[T=\{-3,1\}.\]

For \(P(x):3x+1\le16\) over \(\mathbb R\), solving gives \(3x\le15\), hence \(x\le5\). The truth set is the interval \((-\infty,5]\).

For \(P(x):x^2-5x=-6\) over \(\mathbb Q\), rewrite as \(x^2-5x+6=0\), factor as \((x-2)(x-3)=0\), and keep the rational solutions. The truth set is \(\{2,3\}\).

Predicates With More Than One Variable

A predicate can involve several variables. For variables \(x 1,x 2,\ldots,x n\), the predicate notation is \(P(x 1,x 2,\ldots,x n)\). If the variables come from universal sets \(U 1,U 2,\ldots,U n\), then the combined input space is the Cartesian product \[U=U 1\times U 2\times\cdots\times U n.\]

The truth set is a set of ordered tuples: \[T=\{(x 1,x 2,\ldots,x n)\in U\mid P(x 1,x 2,\ldots,x n)\}.\]