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I have to argue that a given logical expression is satisfiable and not universally applicable. This is quite easy because all I have to do is find for which values is the expression true and false. However, I want to know how to write the solution formally. For example, let's say that we have the following expression:

$$\alpha_i := X \lor \lnot X$$

Due to the the laws of contradiction, we know that the expression is always true. But if I want to say that: "Let $X=T$ , $T$ is shorthand for TRUTH". How can I let $X$ take on the value of TRUTH?

This might be badly explained, but essentially I want to know how to formally write "let $X$ take on the value of TRUTH".

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If you want to assume that the formula takes the value $True$ under some valuation, but may not be always true, write "Let $V(X) := True$", or "$[\![X]\!]_V := True$", where $V$ is the valuation function assigning truth values to formulas. (Depending on your textbook, it may call and symbolize that function differently, but it surely has been introduced somewhere.)

Then, after calculating the truth values of the compound formulas, you can say something like "With $V_1$ such that $V_1(X) = 1$, there exists a valuation under which the formula $X \land Y \to Z$ is true, so the formula is satisfiable, but under $V_2$ with $V_2(X) = \ldots$ the formula is false, so it is not universally valid".

If you want to say that $X$ is a formula which is tautological, i.e. logically equivalent to a statement of the form $A \lor \neg A$, then you can use the symbol $\top$ and write "Let $X := \top$". $\top$ is an atomic formula which always evaluates to $True$.

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  • $\begingroup$ I think in logic, $1$ and $0$ have set meanings ($T/F$) when used with logical statements? $\endgroup$
    – Shubham Johri
    Commented Nov 7, 2020 at 12:57
  • $\begingroup$ @Shubham Johri Yes, but formulas and truth values are different kinds of mathematical objects, so it doesn't make sense to put an equals sign between them. That's like saying $\{a, b, c\} = 3$, or $f = 23$ (where $f$ is $n$-ary function). Some sources may write it that way, but it's bad practice. Instead, one should say that the value of the truth assignment function applied to the formula equals a truth value: $V: FML \to \{0, 1\}$, so $V(X) \in \{0, 1\}$. $\endgroup$
    – Natalie Clarius
    Commented Nov 7, 2020 at 13:02
  • $\begingroup$ I have removed my comment so you should edit your answer... $\endgroup$
    – Shubham Johri
    Commented Nov 7, 2020 at 13:06
  • $\begingroup$ I think the part about "Don't do $X=1/T$..." was relevant since the OP themselves consider doing that. I referred to the phrase "as suggested in the comments". Sorry for the confusion. $\endgroup$
    – Shubham Johri
    Commented Nov 7, 2020 at 13:10
  • $\begingroup$ @Shubham Johri Yes, that was in response to your comment and is no longer needed now, I deleted it. Thanks for taking up on my argument! $\endgroup$
    – Natalie Clarius
    Commented Nov 7, 2020 at 13:11

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