Hilbert Proof System Instance 2 is Consistent

Theorem

Instance 2 of the Hilbert proof systems $\mathscr H_2$ is consistent.

Proof

Consider Instance 1 of a constructed semantics, denoted $\mathscr C_1$.

Note that $\neg p$ is not a tautology for $\mathscr C_1$.

We will establish that every $\mathscr H_2$-theorem is a $\mathscr C_1$-tautology.

That is, that $\mathscr H_2$ is sound for $\mathscr C_2$.

Starting with the axioms:

 $(A1)$ $:$ Rule of Idempotence $\displaystyle (p \lor p) \implies p$ Proof of Tautology $(A2)$ $:$ Rule of Addition $\displaystyle q \implies (p \lor q)$ Proof of Tautology $(A3)$ $:$ Rule of Commutation $\displaystyle (p \lor q) \implies (q \lor p)$ Proof of Tautology $(A4)$ $:$ Factor Principle $\displaystyle (q \implies r) \implies \left({ (p \lor q) \implies (p \lor r)}\right)$ Proof of Tautology

Next it needs to be shown that the rules of inference of $\mathscr H_2$ preserve $\mathscr C_1$-tautologies.

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 $RST \, 1$ preserves $\mathscr C_1$-tautologies.

Rule $RST \, 2$: Rule of Substitution by Definition

Because the definition of $\mathscr C_1$ 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:

 $\displaystyle \mathbf A \land \mathbf B$ $=$ $\displaystyle 2 \land 2$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \neg ( \neg 2 \lor \neg 2 )$ $\quad$ Rule $RST \, 2 \, (1)$ $\quad$ $\displaystyle$ $=$ $\displaystyle \neg ( 1 \lor 1 )$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \neg 1$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 2$ $\quad$ $\quad$

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

Hence $\mathscr H_2$ is sound for $\mathscr C_1$.

In particular:

$\not\vdash_{\mathscr H_2} \neg p$

Hence $\mathscr H_2$ is consistent.

$\blacksquare$