Are all mathematical conclusions based on truth values (example)?

Question

In maths it is assumed that every statement is either true (t) or false (f). When proving theorems such as: For all vector spaces $V$ over a field $K$ and $\lambda \in K$ it holds that $f_{\lambda}:V \to V, v \mapsto \lambda v$ is a linear map. When proving such a statement I usually did not think of truth values at all, the steps in the proof rather felt "intuitive" and were thus correct to me. However, the formal reason why a proof is correct is always connected to truth values isn't it? This would mean that a formal proof would then look like this:

Let $V$ be a vector space over $K$ and $\lambda \in K$ (this means that $V$ is a vector space is a true statement, as well as $K$ is a field and $\lambda \in K$ are true as well). Since $V$ is a vector space, it is by definition true that $V$ is a set. Thus $f_{\lambda} \subseteq V \times V$ with $f_{\lambda}:=\{(v,\lambda v) \ | \ v \in V\}$ is a subset (this is true by set theoretic axioms). Furthermore it holds that if $v_1=v_2$, then $\lambda v_1 = \lambda v_2$ since $V$ is equipped with a well defined multiplication function and by definition of a function this implication is true. Therefore $f_{\lambda}$ is a function. Since $f_{\lambda}$ is a function the element $f_{\lambda}(v_1+v_2)$ exists for all $v_1,v_2 \in V$ and it is true that $f_{\lambda}(v_1+v_2)=f_{\lambda}(v_1)+f_{\lambda}(v_2)$ using the axioms of a vector space which were assumed to be true. Thus the statement is true. Here I am also unsure if it is necessary to know that $f_{\lambda}(v_1+v_2)$ exists in order to proceed in the proof and what harm it would cause if it would not. Then the statement would be somewhat vacuous, I suppose.

In textbooks I don't usually see authors saying that the step in their proof is true, but I am correct in saying that this is what is meant when proving things, right? I suspect that when authors write that "something holds" they also mean that the truth value of the statement is (t). Another question is, whether it is good to think this way. I would say that, when one is insecure whether ones proof is correct, this might be helpful, however it makes proofs a lot longer and restricts the readability to permanently write down that something is true.

Answer 1

It seems to me that, so far you have encountered the (metaphorically speaking) coin that represents logic and proofs but you haven't seen or been shown yet, that it really has two individual sides.

There is as a deductive and a semantic side of it.

On the deductive side, one settles on a set of inference rules which are deemed valid logical principles. For example (where $A, B, C$ are any propositions):

• If $A$ and $A \to B$ then we can conclude $B$.
• If we want to show $C$, we can instead try to show $A \to C$ and $A$.
• If $A$ and $B$ can be shown then $A \land B$.
• If $A \land B$ then we can conclude $A$.

And there are a bunch more. These can then be used to give proofs, and in the context of mathematics these are usually presented in the form of text. Let's write $\vdash P$ when we have given a poof like this for the proposition $P$.

On the semantic side, we proceed differently. We settle to assigning truth values true $T$ or false $F$ to propositions. If the proposition is build from other propositions by combining them via a logical connective like $\to, \lor$ or $\land$, we specify how the truth value of the composite proposition is defined. This is usually presented in the form of truth tables; like here in this case of $\land$ :

$$\boxed{\begin{array}{cc|c} A & B & A \land B \\ \hline T & T & T \\ T & F & F \\ F & T & F \\ F & F & F \end{array}}$$ Using these we can figure out the truth value of any composite proposition $P$ when we are given the truth values of the propositions it is build up from (let's call this the set $\mathsf{atoms}(P)$). If $P$ is true no matter the truth values of the propositions in $\mathsf{atoms}(P)$, then we call $P$ semantically true and let's write $\vDash P$ to express this. This is why you usually do a table consisting of all the propositions in $\mathsf{atoms}(P)$, try out all their possible truth value assignments over several rows, and check if $P$ always gets $\mathsf{true}$ as a result.

And now, in the best case scenario, when we have chosen the right deduction rules and the right definition of semantics, we get a wonderful connection between the two sides of the coin:

$\vdash P ~~\Longleftrightarrow ~~\vDash P$

Or in words: There is a deductive proof of $P$ if and only if it is semantically true. And by this connection there is no need to repeat writing "which is true" or similar statements during a text-style proof, since we are on one side of the coin then.

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To wrap this up, I will give an example by proving the proposition $(A \to B) \lor (B \to C)$ in both ways.

First the more commonly known way, using a truth table:

$$\boxed{\begin{array}{ccc|ccc} A & B & C & A \to B & B \to C & (A \to B) \lor (B \to C) \\ \hline T & T & T & T & T & T \\ T & T & F & T & F & T \\ T & F & T & F & T & T \\ T & F & F & F & T & T \\ F & T & T & T & T & T \\ F & T & F & T & F & T \\ F & F & T & T & T & T \\ F & F & F & T & T & T \end{array}}$$ Next with a deductive text-style proof where I implicitly use the inference rules for classical propositional logic:

We need to show $(A \to B) \lor (B \to C)$. By the law of excluded middle we know that either $B$ or $\neg B$.

• In the case that we have $H_1 : B$ we can show $A \to B$, since assuming $A$, we can already show $B$ because of $H_1$. Because of this we can in particular show $(A \to B) \lor (B \to C)$.
• If we have $H_2 : \neg B$, then we can prove $B \to C$, since given $B$ we have a contradiction to $H_2$ and therefore by explosion (always sounds so dramatic!) we have $C$. And again, this also shows $(A \to B) \lor (B \to C)$. $\hspace{0.3em} \Box$

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If you want a more thourough introduction into these aspects, I can really recommend the books by Peter Smith on Logic. Some are freely available on his website.

Answer 2

Lets take another viewpoint. Assume that you have a certain mathematical theory. Some axioms regarding this theory (e.g. in the theory of sets you would probably take the ZFC axioms... etc. and of course with logical axioms, equalitarian axioms and so on). In this case you have a "statement" of the form "$A\Rightarrow B$" which you want to show(in this theory). Now what you are doing is actually taking a new axiom into your theory (In this case the hypothesis $A$). That is to say you assume $A$ to be "true". So that in the new theory you have obtained by adjoining this new axiom $A$, $B$ is a "true statement". Therefore $A\Rightarrow B$ is a "true statement" in your old theory. So whenever you have a statement of the form "$A\Rightarrow B$" which you want to show is true in your theory, what you are doing is, assuming $A$ to be "true", then with that working in a new theory in which $A$ is added as an axiom to your old theory, showing that $B$ is "true" in this new theory, then you throw away your new theory(you do not need it anymore), because in your old theory $A\Rightarrow B$ is now shown to be "true". Hope that answers your question somewhat.