# Compound propositions as assertions?

According to comments on my previous question, compound propositions are not assertions; i.e. the statement "$p \vee q$" does not mean "$p$ (is true) or $q$ (is true)", and it does not mean "$(p$ or $q)$ is true".From my understanding, "$p \vee q$" is just an un-evaluated expression, just like $2+3$.

However, my book says

The statement $p \implies q$ is called a conditional statement because $p \implies q$ asserts that $q$ is true on the condition that $p$ holds

which seems to say that we can interpret "$p \implies q$" as "$p$ (is true) $\implies q$ (is true)". This appears to directly contradict the comments in the other question, especially since we can write a conditional statement as a disjunction. Indeed, if we assume that "$p \vee q$" means "$p$ (is true) or $q$ (is true)", the truth tables match. If we assume "$p \vee q$" means "$p$ (is false) or $q$ (is false)" instead, the truth table does not match with $p \vee q$; "$p$ (is false) or $q$ (is true)" (or the other way around) doesn't match either.

So my question is:

Is there a contradiction between the comments on my previous question and the book?

If anyone is wondering, my book is "Discrete Mathematics and Its Applications" $6^{th}$ edition by Kenneth H. Rosen.

• What do you mean exactly with "write a conditional statement as a disjunction"? – janezdu Jun 28 '16 at 19:12
• @janezdu I mean that $p \implies q$ is equivalent to $~p \vee q$. My point was that if we assume an assertion of truth only in the case of conditional statements then there wouldn't be a conflict between the book and my other question, but since both refer to disjunctions there appears to be one – Ovi Jun 28 '16 at 19:15
• Those two aren't equivalent; for example, when both $p, q$ are false, $p \implies q$ evaluates to true while $p \vee q$ is false. – janezdu Jun 28 '16 at 19:26
• @janezdu oops i guess my negation symbol didn't register, i meant to write "not $p \vee q$" – Ovi Jun 28 '16 at 19:32
• Why do you think that compound propositions are not assertions ? See Rosen, page 33: "To encode a Sudoku puzzle, let $p(i, j, n)$ denote the proposition that is true when the number $n$ is in the cell in the $i$th row and $j$ th column. There are $9 × 9 × 9 = 729$ such propositions [...]. Given a particular Sudoku puzzle, we begin by encoding each of the given values. Then, we construct compound propositions that assert that every row contains every number, every column contains every number [...] 1/2 – Mauro ALLEGRANZA Jul 2 '16 at 15:49

The conditional statement asserts asserts that $q$ is true on the condition that $q$ holds, but the statement itself can be wrong. I can say that my statement $r$ is "if I never study ($p$), then I will always get 100% on my exams ($q$)". Then, when I refuse to study and fail the exam, $r$ itself is shown to be false.

In other words, $P \implies Q$ tries to assert something, but it is an unevaluated expression as much as $P \vee Q$ is.

I also used the Rosen textbook (7th edition, doesn't matter)

• I understand that the statement itself can be evaluated as false, but in the other question question, David C Ullrich seems to say that we cannot just add (is true) to any proposition. The books seems to say we can. – Ovi Jun 28 '16 at 19:18
• I don't see where Rosen says we can add "is true" arbitrarily? The conditional statement is a proposition because it can be evaluated to true or false given an interpretation. That is, I can say whether $p \implies q$ is true or false if you tell me what $p$ and $q$ are individually (that's the interpretation). – janezdu Jun 28 '16 at 19:29
• @Ovi: David is not, as far as I can see, saying that we cannot add "is true" to things. What he's saying is that there's no need to do so (in ordinary informal mathematical prose), because it doesn't change the meaning being expressed. – Henning Makholm Jun 28 '16 at 19:29
• What I actually said: _Of course when a person states something he's saying that that something is true. That's precisely why we don't need to add "is true" to that statement... _ – David C. Ullrich Jun 28 '16 at 20:39

Recall that if $p$ and $q$ are propositions, the compound proposition $p \vee q$ can be defined by its truth table. This truth table has $4$ rows, and the truth value of $p \vee q$ will be $F$ only in the row where both $p$ and $q$ have truth value $F$.

Similarly, the conditional statement $p \implies q$ is defined to be the proposition $(\sim p) \vee q$ whose truth value is $F$ only when $p$ is $T$ and $q$ is $F$.