Proving well formed propositional statements have well defined truth values

Question

I'm trying to prove by induction that "Any well-formed propositional statement has a well-defined truth value.

The issue is that this seems like a rather mushy thing to prove, so I would also like clarification on what "well-defined means". My intuitive grasp doesn't mesh with the definition (if you can call it that) given in my text:

"In mathematics, when you ask whether something is “well-defined”, you mean that somewhere in the definition a choice was made, and you want to know whether a different choice would have resulted in the same final result. For example, let $X_1 = ({−2,2})$ and let $X_2 = ({−1,2})$. Define $y_1$ by: “Choose x in X1 and let $y_1 = x^2.”$ Define $y_2$ by: “Choose x in $X_2$ and let $y_2 = x^2.”$ Then y1 is well-defined, and is the number 4; but $y_2$ is not well-defined, as different choices of x give rise to different numbers."

My understanding of well-defined seems to be pretty mushy, and seems to be basically injectivity: that a choice of element in the range should have an unambiguous inverse image. I know this isn't quite right, but the above example doesn't help; it seems like $y_2$ is well defined, because you know exactly which element of $X_2$ it comes from. If someone could help me untangle these ideas, it would be helpful.

Here is my attempt at a proof of the proposition above:

Consider a well-formed statement P with definite truth value $a|a\in{0, 1}$. Such a statement clearly exists ("I have a nose", for example).

Now consider a compound statement $Q = Q_1 c_1 Q_2 c_2...Q_n$ where $Q_i$ is a well formed statement, as above, and $c_i$ is a connective with a prescribed truth value. Assume Q has a well defined truth value.

Let $Q' = Q_1 c_1 Q_2 c_2...Q_n c_n Q_{n+1} = Qc_nQ_{n+1}$ Because $Q$ and $Q_{n+1}$ have well defined truth values, and the connective $c_n$ does, so too does $Q'$. Thus, by induction, the proposition is true for all well-formed propositional statements.

Any feedback or clarification of any of the above would be greatly appreciated. Thanks!

Answer 1

So what does this question mean?

Suppose $Q =Q_1c_1Q_2c_2 \ldots c_n Q_n$ is a compound statement. To find its truth value we inductively find the truth value of each $Q_i$ (eventually this will reduce to checking atomic statements which obviously have well-defined truth-values) and then determine the truth value of $Q$ as the connectives $c_i$ dictate. For example if $c= \wedge$ gives $Q_1 c Q_2$ true iff both $Q_1$ and $Q_2$ are true.

To show this procedure always gives the same answer we must consider alternate expressions $Q = Q_1'c_1'Q_2'c_2' \ldots c_m' Q_m'$, apply the above algorithm to the new expression, and prove the answer is the same.

For example $(\neg A) \wedge B \wedge C$ is the same as $(\neg A) \wedge( B \wedge C)$.

Answer 2

'Well-defined' has nothing to do with injectivity; that is where you are confused. 'Well-defined' in this context simply means that every proposition has exactly one truth-value. To show this, you show:

Base: ever atomic proposition has exactly one truth-value (refer to definition of atomic statements and their truth-values here)

Step: Any complex statement that is the result of applying some logical operator to a bunch of other statements has exactly one truth-value assuming (hypothesis) those other statements have exactly one truth-value (refer to truth-functional nature of each defined operator here).

That's all!