Hilbert Proof System Instance 2 is Consistent

Theorem
The axiom system of Principia Mathematica is consistent.

Proof
For the logical connectives Contradiction and Disjunction, let a model be defined as follows.

Let the propositional variables $p, q, r, \dotsc$ be assigned the values $1$ or $2$, according to the following truth tables:


 * $\begin{array}{|c||c|} \hline

& \neg \\ \hline 1 & 2 \\ 2 & 1 \\ \hline \end{array} \qquad \begin{array}{c|cc} \lor & 1 & 2 \\ \hline 1 & 1 & 2 \\ 2 & 2 & 2 \\ \end{array}$

From Functionally Complete Logical Connectives: Negation and Disjunction, $\left\{ {\neg, \lor}\right\}$ is functionally complete.

Thus it is possible to evaluate any WFF and assign its main connective one of the values $1$ or $2$ for each combination of values of its propositional variables.

In particular it is possible to do this on each of the axioms in the axiom system.

Hence as follows:

Axiom $1$: Rule of Idempotence
The Rule of Idempotence:
 * $\left({p \lor p}\right) \implies p$

By Rule of Material Implication, this can be written as:


 * $\neg \left({p \lor p}\right) \lor p$

which evaluates by truth table to:


 * $\begin{array}{|cccc|c|c|} \hline

\neg & (p & \lor & p) & \lor & p \\ \hline 2 & 2 & 1 & 1 & 2 & 1 \\ 1 & 2 & 2 & 2 & 2 & 2 \\ \hline \end{array}$

Axiom $2$: Rule of Addition
The Rule of Addition:
 * $q \implies \left({p \lor q}\right)$

By Rule of Material Implication, this can be written as:


 * $\neg q \lor \left({p \lor q}\right)$

which evaluates by truth table to:


 * $\begin{array}{|cc|c|ccc|} \hline

\neg & q & \lor & (p & \lor & q) \\ \hline 2 & 1 & 2 & 1 & 1 & 1 \\ 1 & 2 & 2 & 1 & 2 & 2 \\ 2 & 1 & 2 & 2 & 2 & 1 \\ 1 & 2 & 2 & 2 & 2 & 2 \\ \hline \end{array}$

Axiom $3$: Rule of Commutation
The Rule of Commutation:
 * $\left({p \lor q}\right) \implies \left({q \lor p}\right)$

By Rule of Material Implication, this can be written as:


 * $\neg \left({p \lor q}\right) \lor \left({q \lor p}\right)$

which evaluates by truth table to:


 * $\begin{array}{|cccc|c|ccc|} \hline

\neg & (p & \lor & q) & \lor & (q & \lor & p) \\ \hline 2 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 2 & 2 & 2 & 2 & 2 & 1 \\ 1 & 2 & 2 & 1 & 2 & 1 & 2 & 2 \\ 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ \hline \end{array}$

Axiom $4$: Factor Principle
The Factor Principle:
 * $\left({p \implies q}\right) \implies \left({\left({r \lor p}\right) \implies \left ({r \lor q}\right)}\right)$

By Rule of Material Implication, this can be written as:


 * $\neg \left({\neg p \lor q}\right) \lor \left({\neg \left({r \lor p}\right) \lor \left ({r \lor q}\right)}\right)$

which evaluates by truth table to:


 * $\begin{array}{|ccccc|c|cccccccc|} \hline

\neg & (\neg & p & \lor & q) & \lor & (\neg & (r & \lor & p) & \lor & (r & \lor & q)) \\ \hline 1 & 2 & 1 & 2 & 1 & 2 & 2 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 2 & 1 & 2 & 2 & 2 & 1 \\ 1 & 2 & 1 & 2 & 2 & 2 & 2 & 1 & 1 & 1 & 2 & 1 & 2 & 2 \\ 1 & 2 & 1 & 2 & 2 & 2 & 1 & 2 & 2 & 1 & 2 & 2 & 2 & 2 \\ 2 & 1 & 2 & 1 & 1 & 2 & 1 & 1 & 2 & 2 & 1 & 1 & 1 & 1 \\ 2 & 1 & 2 & 1 & 1 & 2 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 1 \\ 1 & 1 & 2 & 2 & 2 & 2 & 1 & 1 & 2 & 2 & 2 & 1 & 2 & 2 \\ 1 & 1 & 2 & 2 & 2 & 2 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ \hline \end{array}$

As can be seen, the main connective of each of the axioms evaluates to $2$.

Next it needs to be shown that this property is retained when these axioms have the proof rules applied to them.