Hilbert Proof System Instance 2 is Consistent

Theorem
Instance 2 of the Hilbert proof systems is consistent.

Proof
For the logical connectives Negation and Disjunction, let a formal semantics be defined as follows.

Let the propositional variables $p, q, r, \dotsc$ be assigned the values $1$ or $2$.

Next, let the assignment be extended to the WFFs according to the following 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}$

By the definitional abbreviations of Rule $RST \, 2$, 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 system.

Hence as follows:

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

By Rule $RST \, 2$, definition $(2)$, this can be written as:


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

which evaluates 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 $RST \, 2$, definition $(2)$, this can be written as:


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

which evaluates 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 $RST \, 2$, definition $(2)$, this can be written as:


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

which evaluates 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 $RST \, 2$, definition $(2)$, 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 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 rules of inference of the system applied to them.

Rule $RST \, 1$: Rule of Uniform Substitution
By definition, any WFF is assigned a value $1$ or $2$.

Thus, in applying Rule $RST \, 1$, we are introducing $1$ or $2$ in the position of a propositional variable.

But all possibilities of assignments of $1$s and $2$s to such propositional variables were shown not to affect the resulting value $2$ of the axioms.

Hence Rule $RST \, 1$ takes expressions with value only $2$ to expressions with value only $2$.

Rule $RST \, 2$: Rule of Substitution by Definition
Because the definition of the formal semantics was given in terms of Rule $RST \, 2$, it cannot affect any of its results.

Rule $RST \, 3$: Rule of Detachment
Suppose $\mathbf A$ and $\mathbf A \implies \mathbf B$ both take value $2$.

Then using Rule $RST \, 2$, definition $(2)$, we get:


 * $\neg \mathbf A \lor \mathbf B$

taking value $2$ by assumption.

But $\neg \mathbf A$ takes value $1$ by definition of $\neg$.

So from the definition of $\lor$ it must be that $\mathbf B$ takes value $2$.

Hence Rule $RST \, 3$ also produces only WFFs of value $2$.

Rule $RST \, 4$: Rule of Adjunction
Suppose $\mathbf A$ and $\mathbf B$ take value $2$.

Then:

proving that Rule $RST \, 4$ also produces only $2$s from $2$s.

Hence no application of axioms and rules can result in derivation of an expression that takes only the value $1$.

That is, Instance 2 is consistent