# Can $p ∧ q$, where $p$ and $q$ are propositional variables, be a proposition?

Edit:
I have removed the original text in favor of showing the specific passages in the book that I am stuck on. Sorry about the inconvenience.

I am wondering whether $$p ∧ q$$ can be a proposition when $$p$$ and $$q$$ are both propositional variables. It seems to me that my book in one case says that a proposition can't have a variable truth value and then states that $$p ∧ q$$ is a proposition, even though it surely also has a variable truth value.

Page 2:

Consider the following sentences.

1. What time is it?
3. $$x + 1 = 2$$
4. $$x + y = z$$

Sentences 1 and 2 are not propositions because they are not declarative sentences. Sentences 3 and 4 are not propositions because they are neither true nor false. Note that each of sentences 3 and 4 can be turned into a proposition if we assign values to the variables.

Page 26:

Note that we will use the term "compound proposition" to refer to an expression formed from propositional variables using logical operators, such as $$p ∧ q$$.

Rosen, K. H. (2019). Discrete Mathematics and Its Applications Eighth Edition. New York, NY: McGraw-Hill Education.

Couldn't we have stated the following?

"$$p ∧ q$$ is not a proposition because it is neither true nor false. Note that it could be turned into a proposition if we assign values to the variables $$p$$ and $$q$$."

• Surely that would come under the “fair use” exception? – Lubin Sep 4 '19 at 20:01
• Here, you are using it for scholarly analysis, and the copyright infringement contains this as an exception. or else you can quote the page number and the name of the book. – Kumar Sep 4 '19 at 20:41
• $x=5$ here should be interpreted as "$x$ is defined to be 5". The way you're thinking about that is not wrong per se, but it is not the intended interpretation in this case. Some authors use $:=$ to denote a definition or even $\equiv$ to distinguish definition from equality. – Cameron Williams Sep 5 '19 at 10:34
• @CameronWilliams The statement $x = 5$ is my rewriting of the original $x + 1 = 2$. There it is made clear that x is a variable. Therefore, the statement isn't meant to define x as a given constant value. – Name Sep 5 '19 at 11:16
• @CameronWilliams I have edited the post now. Hopefully it is more clear. – Name Sep 5 '19 at 11:46

$$x=5$$ is said to be a predicate of one variable denoted by $$P(x)$$.

$$p \land q$$ can also be seen as a predicate (of two variables) whose truth value (val) depends on truth values of $$p$$ and $$q$$.

We could denote it by $$Q(val(p),val(q))$$ for example

$$val(Q(1,0))=0$$

If $$x$$ is a numerical constant, $$x=5$$ is a proposition; otherwise, it can be thought of as a number-to-boolean function. If $$p,\,q$$ are boolean or propositional constants, $$p\land q$$ is a proposition; otherwise, it is a truth function.

Some of this depends on how one's textbook defines these formulas.

According to the forallx logic textbook, "the symbol ‘=’ is a two-place predicate" in first-order logic with identity. The formula $$x=5$$ asserts that $$x$$ "is identical to" $$5$$. Both $$x$$ and $$5$$ "name the same thing". (page 205) It may not make sense to think of this formula as true or false.

Rather than calling $$p$$ and $$q$$ 'propositions', this textbook calls them 'atomic sentences' and 'subsentences' of the sentence $$p\land q$$ in truth-functional logic. (page 23) One can assign truth values to these 'sentence letters'. One could also refer to these as propositions in truth-functional logic although this textbook doesn't use that term.

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf

I think the right way to think about it is this: If you just say $$p \wedge q$$, with no context, then that's right, it is not a proposition, because it is just a string of symbols withouth meaning, so it is neither true or false. However, if you give the context "p is a proposition and "q is a proposition", then "$$p \wedge q$$" is definitely a proposition.

Similarly with $$x=5.$$ With no context, this is just a string of sybols with no meaning. If you give the context "$x is an irrational number", then this becomes a proposition (a false one). If you give the context "x is 5", then this is again a proposition, a true one. • "However, if you give the context "p is a proposition and "q is a proposition", then "p∧q" is definitely a proposition." I agree. What confuses me is that$p$and$q$are propositional variables, not propositions. I added the excerpts from the book also. Thanks for the help. – Name Sep 5 '19 at 11:23 • @Name Hmm then in that case I would just go with what the book says. I am not a logic person. If it makes you feel any better, unless you’re studying logic, the answer to this question won’t affect anything. – Ovi Sep 5 '19 at 14:35 It seems to me that you are mix-and-matching different contexts ... The first quote is from page 2, and the author plainly states the basic concept of proposition : A proposition is a declarative sentence (that is, a sentence that declares a fact) that is either true or false, but not both. Here the context is natural language and Examples 1&2 give natural language sentences : some are declarative sentences, stating a fact that is either true or false (e.g. Toronto is the capital of Canada) while Others are not (e.g. What time is it? : it is a question and does not state a fact). Immediately after, the author "moves to" the preliminaries of propostional calculus saying : We use letters to denote propositional variables (or statement variables), that is, variables that represent propositions, just as letters are used to denote numerical variables. This means that, when the author writes : Let $$p$$ be a proposition. The negation of $$p$$, denoted by $$¬p$$, is the statement “It is not the case that $$p$$.” The proposition $$¬p$$ is read “not $$p$$.” he is stating the first rule of the syntax of the calculus, concerning the negation sign. How we have to interpret it ? In this way : every time we replace the statement variable $$p$$ in the formula $$\lnot p$$ with a declarative sentence of natural Language, what we get is a new declarative sentence that is the negation of the first one. Thus, let $$p$$ be replaced by the sentence "Toronto is the capital of Canada"; the corersponding formula $$\lnot p$$ will be : "Toronto is not the capital of Canada". The same for the conjunction connective [page 4] : Let $$p$$ and $$q$$ be propositions. The conjunction of $$p$$ and $$q$$, denoted by [the formula] $$p ∧ q$$, is the proposition “$$p$$ and $$q$$.” In conclusion, regarding the title question : Can $$p ∧ q$$, where $$p$$ and $$q$$ are propositional variables, be a proposition? the answer is : $$p ∧ q$$ is a formula that will become a proposition when the propositional variables $$p,q$$ will be replaced by propositions. Note : consider again the definition above, where the author says : just as letters are used to denote numerical variables. When we write "the number $$n+1$$..." we are not saying that $$n$$ is a number, but we are saying that whatever number we assign as value to the numerical variable $$n$$ the result will be a number. Regarding the fact that the arithmetical formula $$x+1=2$$ is not a declarative sentence in the language of arithmetic, you can see the post Is logical implication always determinable from just the given statements for a discussion of similar examples. • I think I have come to a partial understanding by considering functions, which present an analogous situation. Say we have a function$f(x) = |x|$. Then we could talk about "the function value$f(x)$", even though it isn't a function value unless you specify$x$. Similarly, we can talk about "the proposition$p ∧ q$" even though it isn't a proposition unless you specify$p$and$q$. So in conclusion, it is simply a matter of phrasing or terminology. Why$x + 1 = 2$doesn't have a truth value is something I'll have to look into. I'll check your link when I have the time. – Name Sep 5 '19 at 18:19 • Would you say that's fair or am I missing something? – Name Sep 5 '19 at 18:24 • @Name - correct; in the context of propositional calculus we can talk about "the proposition$p∧q$" even though it isn't a proposition unless we specify$p$and$q$. In other words, the "rule of the game" is that whenever$p,q$are specified, the compound formula$p \land q\$ will be a proposition also. – Mauro ALLEGRANZA Sep 6 '19 at 6:14
• Is there any obvious reason why this phrasing is allowed? To me, it seems imprecise. – Name Sep 6 '19 at 19:18