Working on the book: Derek Goldrei. "Propositional and Predicate Calculus" (p. 40)

The author briefly introduces the concept of a function v:

We talked informally about knowing whether, in a particular set of circumstances, each propositional variable is true or false. More formally and elegantly, thsi set of circumstances is a function $v \colon P \to \{T,F\}$, where $P$ is the set of propositional variables in our language.

Before introducing the concept of truth assignment, explains:

Let $Form(P, S)$ be the set of all formulas built up from propositional variables in a set $P$ using connectives in a set $S$ which includes $\land$. We shall say that a function $v \colon Form(P,S) \to \{T,F\}$ respects the truth table $\land$ if $$ v((\theta \land \psi))= \begin{cases} T, &\text{if } v(\theta))=v(\psi)=T\\ F, &otherwise, \\ \end{cases} $$ for all formulas $\theta, \psi \in Form(P,S)$.

I see how function $v$ respects the truth table of $\land$. My question is:

  • In which way are the truth value of $v(\theta)$ and $v(\psi)$ determined in that piecewise function ?
  • What is the criteria for deciding whether $v(\theta)$ (or any propositional variable) is going to be $T$ or $F$ ?
  • $\begingroup$ See van Dalen, page 17 for the definition and the basic result: "If a valuation is only given for atoms then it is, by virtue of the definition by recursion [see your previous post], possible to extend it to all propositions. " $\endgroup$ Oct 8, 2020 at 6:40
  • $\begingroup$ Thank you so much, @Mauro ALLEGRANZA. That's exactly what I do not understand. Could you explain a little bit ? What does "...possible to extend it to all propositions", mean ? I see how the value of compound formulas is computes; I just do not see it when it comes to atoms. I mean, what determines the assignment of T or F to an atomic formula. $\endgroup$
    – F. Zer
    Oct 8, 2020 at 12:05
  • $\begingroup$ Thank for the offering. I will appreciate if you could clarify the meaning of "...possible to extend it to all propositions". Also, is perhaps the assignment to atoms random ? I feel this is not really complicated but a detail evades me. $\endgroup$
    – F. Zer
    Oct 8, 2020 at 12:32

3 Answers 3


Ultimately the truth value $v(\varphi)$ of any formula $\varphi$ is determined by the truth values that $v$ assigns to the propositional variable appearing in $\varphi$. Consider, for instance the following formula $\varphi$:

$$(p\land\neg q)\lor(r\land q)\to p$$

If $v(p)=F$, $v(q)=T$, and $v(r)=F$, then:

  • $v(\neg q)$ must be $F$ in order to respect to truth table for $\neg$;
  • $v(p\land\neg q)$ must then also be $F$ in order to respect the truth table for $\land$;
  • $v(r\land q)$ must be $F$ for the same reason;
  • $v\big((p\land\neg q)\lor(r\land q)\big)$ must be $F$ in order to respect the truth table for $\lor$; and finally
  • $v(\varphi)$ must be $T$ in order to respect the truth table for $\to$.

If we change the truth value of $r$ by setting $v(r)=T$, we must then have $v(r\land q)=T$, $v\big((p\land\neg q)\lor(r\land q)\big)=T$, and $v(\varphi)=F$.

In practice truth assignments are very often presented in the form of tables showing the truth values assigned to a particular formula under all possible assignments of truth values to its propositional variables. In this case we’d have the following table, and we could read off that $v(\varphi)=T$ for all truth assignments to $p,q$, and $r$ except the one mentioned immediately above, the one in the fifth line of the table:

$$\begin{array}{c|c|c} p&q&r&p\land\neg q&r\land q&(p\land\neg q)\lor(r\land q)&\varphi\\\hline T&T&T&F&T&T&T\\ T&T&F&F&F&F&T\\ T&F&T&T&F&T&T\\ T&F&F&T&F&T&T\\ F&T&T&F&T&T&F\\ F&T&F&F&F&F&T\\ F&F&T&F&F&F&T\\ F&F&F&F&F&F&T \end{array}$$

  • $\begingroup$ Very clear. Thank you so much, @Brian M. Scott ! I see what you explain about respecting the truth table of logical connectives. But, in which way the value of $v(\theta)$ is determined ($\theta$ being an atomic proposition) ? $\endgroup$
    – F. Zer
    Oct 8, 2020 at 11:59
  • $\begingroup$ Taking a simple example, I know the value of $\sqrt{4}$ is going to be $2$, but in terms of truth assignments, I cannot see a predictable way of determining if a certain proposition is going to be true or false. $\endgroup$
    – F. Zer
    Oct 8, 2020 at 11:59
  • $\begingroup$ Perhaps, there is a subtle point about functions that I'm not getting... $\endgroup$
    – F. Zer
    Oct 8, 2020 at 12:00
  • $\begingroup$ @F.Zer: I’m not quite sure what you’re asking. Are you asking how, if $p$ is a propositional variable, one determines the truth value of $p$? That’s not a question with which propositional logic deals: propositional logic deals with how truth values of propositions are correctly combined. $\endgroup$ Oct 8, 2020 at 16:27
  • 1
    $\begingroup$ @F.Zer: Yes, that’s correct. For instance, the truth table in my answer shows every one of the $8$ possible truth assignments to the three variable $p,q$, and $r$, one per line. $\endgroup$ Oct 8, 2020 at 17:05

The above answers are thorough, but they overcomplicate matters by quite a bit by being too formal. A truth assignment ν is simply a row of values of atomic propositions in a truth table, nothing more, nothing less.

This row of atomic proposition values can be thought of as a function ν(φ) → {T, F}, which is true or T, if the value of the formula φ is true with that combination of values of atomic propositions ν and false or F, if the fomula φ is not true with said values. For example, if it is raining and cloudy or raining = T and cloudy = T, then

raining ∧ cloudy = T ∧ T = T ,

and the sequence raining cloudy = T T = ν for which ν(φ) = ν(raining ∧ cloudy) = T.

Another way of writing the fact that ν(φ) = T, or that φ is true with the observed combination of values of the atomic propositions, is ν ⊧ φ, or ν is a model of the formula φ. If φ is not true with the given combination of atomic proposition values ν, we say that ν ̸⊧ φ, or that ν is a counter model of φ.

Anyways, thinking about this in terms of truth tables is in my opinion more illustrative than most alternative presentations.


See Dirk van Dalen, Logic and Structure (5th ed., 2013), page 17 for the definition and the basic result:

"If a valuation is only given for atoms then it is, by virtue of the definition by recursion, possible to extend it to all propositions."

What is a valuation ? Simply a function: $v : \text {PROP} \to \{ 0,1 \}$,

where [see page 7] $\text {PROP} = \{ p_0, p_1, \ldots \}$ is the collection of proposition symbols.

Side condition: van Dalen uses $\bot$ and consider it as a $0$-ary connective. Thus, he needs the clause: for every valuation $v$ we must have: $v(\bot)=0$.

Silly examples of valuations:

$v(p_i)=0$, for every $i$; $v'(p_i)=1$, for every $i$.

Starting with an assignment $v$, we can easily show how the truth tables for the basic connectives [see page 18] give us the recipe for computing the truth value for a formula $\varphi$ whatever.

Consider the formula $((p_0 ∧ p_1) → (\lnot p_0))$ [similar to the example used in your previous post] and build the usual truth table:

$$\begin{array}{c|c|c} p_0&p_1&(p_0 \land p_1)&(\lnot p_0)&((p_0\land p_1)\to(\lnot p_0))&\\\hline 1&1&1&0&0\\ 1&0&0&0&1\\ 0&1&0&1&1\\ 0&0&0&1&1\\ \end{array}$$

Consider now the following valuation: $v(p_0)=0$ and $v(p_1)=1$. It simply amounts to the third line in the table above.

Note: as we can imagine [but we can prove it: see Lemma 2.2.3, page 18], what matters are only the value that the valuation assigns to the propositional symbols occurring into the formula.

The "recipe" is:

(i) $v(p_0)=0$; thus $v[(\lnot p_0)]=1$.

(ii) $v(p_0)=0$ and $v(p_1)=1$; thus $v[(p_0 \land p_1)]=0$.

(iii) $v[(p_0 \land p_1)]=0$ and $v[(\lnot p_0)]=1$; thus $v[((p_0 \land p_1)\to (\lnot p_0))]=1$.

Note: in computing the truth value of the formula, we have followed the formation sequence [see Definition 2.1.4, page 9] of the formula:

$p_0, p_1, (p_0 \land p_1), (\lnot p_0), ((p_0 \land p_1)\to (\lnot p_0))$.


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