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In the last section, we introduced the idea of assigning truth values to statements. This idea is unproblematic for statements that do not contain variables, since such statements are either true or false. But if a statement contains variables, we can no longer describe the statement as being simply true or false. Its truth value might depend on the values of the variables involved. For example, if P(x) stands for the statement “x is a prime number,” then P(x) would be true if x = 23, but false if x = 22. To deal with this complication, we will define truth sets for statements containing variables.

This question might seem pretty basic but forgive my ignorance. When I was being taught truth sets it sounded like we use them to deal with predicates that have variables because we can't simply say they are true or false since it depends on the variable, which can change and therefore affect the validity of the statement. How exactly do truth sets help with this? don't I still need to know what the variable is to determine if the statement is true or false?

numq
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  • Truth sets are found only after doing the work of finding out for which value(s) of the variable(s) the given statement with variables is true. – coffeemath Oct 27 '21 at 06:40
  • The "truth set" is the set of "objects" satisfying predicate $P(x)$. With domain the natural numbers and predicate "x is Even" the corresponding truth set will be ${ 2,4,6, \ldots }$ – Mauro ALLEGRANZA Oct 27 '21 at 06:40
  • Obviously, to determine the truth set we have to identify a domain and an interpretation for the non-logical symbols, like the predicate letter $P$ in the example above. – Mauro ALLEGRANZA Oct 27 '21 at 06:41
  • See these ref 1, 2 – cinch Oct 27 '21 at 06:44

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Think of a predicate's “truth set” as abstractly defining the predicate, i.e., which variable/tuple values the predicate is true for. For example,

  • let the domain of discourse be $\{0,1\},$ and let $P(x,y)$ be a predicate with “truth set” $\{ (0,0), (1,1) \}.\quad$ (So, $P(0,1)$ and $P(1,0)$ are both false.)

    Under this interpretation, $$∀y∃xP(x,y)→∃x∀yP(x,y)$$ must be a false statement.

ryang
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