I am reading Classical Mathematical Logic by Epstein. I am a little bit unclear on some of his terminology:

No [abbreviated] wff such as $p_0 \wedge \neg p_1$ is true or false; that is the form of a proposition, not a proposition. (Page $11$)

A formal wff is true oe false in the model according to whether its realization is true or false. (Page $17$)

If we are given a certain realization and its valuation:

$p_0 =$ the sky is blue (True), $p_1 = $ iron is red (False)

Is $((p_0) \wedge (\neg (p_1)))$ a proposition? Or is it just a statement that happens to be false (as per the second quote) , but it is not a proposition?

Say, the variables are realized, but they are not valuated. Is $p_0 \wedge \neg p_1$ a proposition then?


Epstein defined as follows:

Proposition: "A Written or uttered declerative sentence used in such a way that it is true or false, but not both."

Realization: "A realization is an assignment of propositions to some or all of the propositional variables. The realization of a formal wff is the formula we get when we replace the propositional variables appearing in the formal wf with the propositions assigned to them; it is a semi-formal wff. The semi-formal language for that realization is the collection of realizations of formal wffs all of whose propositional variables are realized. "

Valuation: " We agree that sentences assigned to propositional variables...are propositions... But it is a further agreement to say which thruth-value it has. When we do that we call all the assignments of specific truth-values together a propositional valuation, $v$, and write $v(p) = T$ or $v(p) = F$ according to whether the atomic proposition $p$ is taken to be true or false.

  • $\begingroup$ Wait, iron's not red? en.wikipedia.org/wiki/Iron(III)_oxide $\endgroup$ – user4894 Aug 6 '16 at 3:05
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    $\begingroup$ @user4894 Your knowledge about iron might be a little rusty. $\endgroup$ – Andreas Blass Aug 6 '16 at 4:05
  • $\begingroup$ Look, this is a badly written book! The more you ask about it, the worse it seems to be! Why don't you just try some other book? If you want to learn mathematical logic later, you're probably going to have to unlearn a lot of this silly stuff from Epstein. In mathematics "$p_0 \land \neg p_1$" is in fact a proposition. $\endgroup$ – user21820 Aug 7 '16 at 9:51
  • $\begingroup$ @user21820 I am reading Patrick Suppes at the same time, but I am still reading this book because they seem to complement each other well. For example, I'm in the same spot in both books (defining $\iff$). So far Patrick has given nice intuitive explanations, while Epstein focused on more formal aspects such as defining a formal language $L(\vee, \wedge, \neg, \rightarrow, p_0, p_1,...)$ , proving that each wff can be read in only one way, and more rigorous definitions in general. When Epstein says that $p_0 \wedge \neg p_1$ is not a proposition, he means it before the variables are assigned $\endgroup$ – Ovi Aug 7 '16 at 13:31
  • $\begingroup$ any meaning. I think he does consider $p_0 \wedge \neg p_1$ as propositions once meaning and valuation is assigned, but that's what I'm trying to find out, is $p_0 \wedge \neg p_1$ a proposition as soon as the variables are realized, or do they have to be valuated as well before $p_0 \wedge \neg p_1$ can be called a proposition $\endgroup$ – Ovi Aug 7 '16 at 13:33

From the definitions given (which are not formal enough to be precise), I guess that he considers any well-formed formula made from connecting atomic propositions as a proposition too.

If $p_0,p_1$ are formulae then "$p_0 \land \neg p_1$" is also a formula.

If $p_0,p_1$ are propositions then "$p_0 \land \neg p_1$" is also a proposition.

If $p_0,p_1$ are true in some world then "$p_0 \land \neg p_1$" is false in that world.

The second point means that the string "$p_0 \land \neg p_1$" is not a proposition, but the string "( the earth is flat ) $\land$ $\neg$ ( the moon is made of cheese )" is a proposition but still has no meaning or truth value whatsoever until it is interpreted in some world.

Note that this is my best guess and there may be deviation from the intended meaning of Epstein. Normally in formal logic we do not have the second level because it's simply a second language that is essentially no different from the first (propositional language). After all, we can directly interpret the string "$p_0 \land \neg p_1$" in a world simply by specifying the interpretation of $p_0,p_1$ in that world.


$p_0\wedge\neg p_1$ is not a proposition, merely a construction of logical operators and variables, as by the first quote. The realisation (sky blue and iron not red) turns this construction into a statement, but it is neither true nor false, as at this point we do not have the truth value the expression should evaluate to. The original expression is also neither true nor false, by the second quote.

Such wffs get turned into propositions by equating them with a single truth value: $p_0\wedge\neg p_1=0$, $p_0\wedge\neg p_1=1$ ("sky blue and iron not red" is true/false). The realisations of these are statements that can be assigned true/false values: if I am on Earth, the sky is indeed blue and iron not red, so $p_0\wedge\neg p_1=1$ is a true proposition, and its realisation a true statement. If I was on Titan, where the atmosphere is orange, $p_0\wedge\neg p_1=1$ would be a false proposition and its realisation a false statement.

  • $\begingroup$ So from what I understand, before we realize any variables $p_0 \wedge \neg p_1$ is not a proposition, but after we realize the variables and valuate them as either T or F, they become propositions. What about in the intermediate step? Say we realized the variables, but we haven't assigned truth values to any of them. Is $p_0 \wedge \neg p_1$ a proposition at that point? $\endgroup$ – Ovi Aug 7 '16 at 13:26
  • $\begingroup$ Propositions require an equals sign, a claim that the set of logical operators and variables is true or false. Even if $p_0\wedge\neg p_1$ is realised, it's still not a proposition because there is no equals sign (claim that the statement is true/false). $\endgroup$ – Parcly Taxel Aug 7 '16 at 13:34
  • $\begingroup$ Hm the author seems to imply what you said in the definition of a proposition 'a written or uttered declarative sentence that is used in such a way that it is true or false, but not both' ('used' is the key word here I think) but he does not talk about an equal sign anywhere. He considers 'Ralph is a dog' a propositions, and says nothing of the sort 'Ralph is a dog=T' for example $\endgroup$ – Ovi Aug 7 '16 at 13:51

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