Formalism

Formalism is a philosophy of mathematics, which holds that math is just a game of formal symbol manipulation, and that mathematical objects per se don't exist at all.

Formalism has the same flavor as subjectivism.

Many formalists are only formalists with respect to some parts of math. This includes most working mathematicians, insofar as they hold formalist premises. For example, Hilbert held that finite math is meaningful, but infinite math is just a game of formal symbol manipulation.

For a good example of explicit formalism, see the following quote from Russel and Whitehead about definitions

Definitions are meaningless BS, blah blah blah [TODO not the real quote ;) ]

Logicism

Logicism is a specific type of formalism, which believes that the rules and concepts of the game are reducible to those of logic (sets, set membership, modus ponens, etc). Logic here refers to logic as understood by Frege, rather than logic as understood by e.g. Aristotle.

For example, Logicism defines the natural numbers to be set-theoretic objects like the following

• $0:=\emptyset$
• $1:=0\cup \{0\}=\{\emptyset \}$
• $2:=1\cup \{1\}=\{\emptyset ,\{\emptyset \}\}$
• $3:=2\cup \{2\}=\{\emptyset ,\{\emptyset \},\{\emptyset ,\{\emptyset \}\}\}$

etc.

Examples

Here I will put examples of working mathematicians saying things based on a formalist premise.

The working mathematician almost always has some formalist premises. To some extent, he treats some pieces of math like a game of formal symbol manipulation. However, it is difficult to identify if, in any specific instance, a mathematician is being a formalist. The reason for this is that what is understood by one man a conceptual argument, could be understood by another man as mere symbol manipulation.

The following example is from Gamelin and Greene, Introduction to Topology:

3.1 Theorem: Let $X$ and $Y$ be topological spaces. A function $f:X\rightarrow Y$ is continuous if and only if $f$ is continuous at every point of $X$.

Proof: Suppose that $f$ is continuous at every point of $X$ and that $V$ is open in $Y$. Let $x\in f^{-1}(V)$. Since $f$ is continuous at $x$, there is an open set $U_{x}$ containing $x$ such that $U_{x}\subset f^{-1}(V)$, or $f(U_{x})\subset V$. Set $U=\cup \{U_{x}:x\in f^{-1}(V)\}$. Then $U$ is open and $f(U)\subset V$, so that $U\subset f^{-1}(V)$. Since $U$ includes each point of $f^{-1}(V)$, $U$ coincides with $f^{-1}(V)$ and $f^{-1}(V)$ is open. Thus $f$ is continuous.

Conversely, suppose that $f$ is continuous. Let $x\in X$ and let $V$ be an open set containing $f(x)$. Then $U=f^{-1}(V)$ is an open set containing $x$ and $f(U)\subset V$. Consequently $f$ is continuous at $x$ for all $x\in X$. $\square$

The reason why I say that this proof is formalistic is that it could be easily understood by someone who knows the formal definitions of the objects involved in the proof, but has no conceptual understanding of what it means for a map to be continuous. Indeed, that situation is the case for many students who are learning about topology for the first time.

Whenever you have a proof or a definition that is super formal, that uses the technicalities of the definitions rather than explanations, there's probably a formalist definition lurking around somewhere.

Refutations

Mathematicians usually don't actually do deductions.