Hilbert Proof System Instance 2 Independence Results/Independence of A4

Theorem
Let $\mathscr H_2$ be Instance 2 of the Hilbert proof systems.

Then:

Axiom $(A4)$ is independent from $(A1)$, $(A2)$, $(A3)$.

Proof
Denote with $\mathscr H_2 - (A4)$ the proof system resulting from $\mathscr H_2$ by removing axiom $(A4)$.

Consider $\mathscr C_5$, Instance 5 of constructed semantics.

We will prove that:


 * $\mathscr H_2 - (A4)$ is sound for $\mathscr C_5$;
 * Axiom $(A4)$ is not a tautology in $\mathscr C_5$

which leads to the conclusion that $(A4)$ is not a theorem of $\mathscr H_2 - (A4)$.

Soundness of $\mathscr H_2 - (A4)$ for $\mathscr C_5$
Starting with the axioms:

Next it needs to be shown that the rules of inference of $\mathscr H_2$ preserve $\mathscr C_5$-tautologies.

Rule $RST \, 1$: Rule of Uniform Substitution
By definition, any WFF is assigned a value from $\set{ 0,1,2,3 }$.

Thus, in applying Rule $RST \, 1$, we are introducing $0$, $1$, $2$ or $3$ in the position of a propositional variable.

But all possibilities of assignments to such propositional variables were shown not to affect the resulting values of the axioms.

Hence Rule $RST \, 1$ preserves $\mathscr C_5$-tautologies.

Rule $RST \, 2$: Rule of Substitution by Definition
Because the definition of $\mathscr C_5$ 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 $0$.

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


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

taking value $0$ 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 $0$.

Hence Rule $RST \, 3$ also produces only WFFs of value $0$.

Rule $RST \, 4$: Rule of Adjunction
Suppose $\mathbf A$ and $\mathbf B$ take value $0$.

Then using the definitional abbreviations:


 * $\mathbf A \land \mathbf B =_{\text{def}} \neg ( \neg \mathbf A \lor \neg \mathbf B )$

We compute:

proving that Rule $RST \, 4$ also produces only $0$s from $0$s.

Hence $\mathscr H_2 - (A4)$ is sound for $\mathscr C_5$.

$(A4)$ is not a $\mathscr C_5$-tautology
Recall axiom $(A4)$, the Factor Principle:


 * $(p \lor q) \implies (q \lor p)$

Under $\mathscr C_5$, we can use definitional abbreviations to arrive at:


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

Applying the definition of $\mathscr C_5$, we have the following:


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

\neg & (\neg & p & \lor & q) & \lor & (\neg & (r & \lor & p) & \lor & (r & \lor & q)) \\ \hline 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 3 & 2 & 0 & 0 & 0 & 1 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 1 & 0 & 3 & 0 & 0 & 0 & 1 & 0 & 0 & 3 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 3 & 2 & 0 & 0 & 0 & 3 & 1 & 2 & 2 & 0 & 1 & 0 & 0 \\ 1 & 0 & 3 & 0 & 0 & 0 & 0 & 1 & 3 & 3 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 & 2 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 3 & 2 & 2 & 1 & 0 & 2 & 0 & 0 \\ 1 & 3 & 2 & 0 & 0 & 0 & 3 & 2 & 2 & 2 & 0 & 2 & 0 & 0 \\ 1 & 0 & 3 & 0 & 0 & 0 & 1 & 2 & 0 & 3 & 0 & 2 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 3 & 0 & 0 & 0 & 3 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 3 & 3 & 1 & 0 & 3 & 0 & 0 \\ 1 & 3 & 2 & 0 & 0 & 0 & 1 & 3 & 0 & 2 & 0 & 3 & 0 & 0 \\ 1 & 0 & 3 & 0 & 0 & 0 & 0 & 3 & 3 & 3 & 0 & 3 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 3 & 2 & 3 & 1 & 0 & 1 & 0 & 0 & 2 & 0 & 0 & 0 & 1 \\ 1 & 0 & 3 & 0 & 1 & 0 & 1 & 0 & 0 & 3 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 3 & 2 & 3 & 1 & 0 & 3 & 1 & 2 & 2 & 3 & 1 & 1 & 1 \\ 1 & 0 & 3 & 0 & 1 & 0 & 0 & 1 & 3 & 3 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 2 & 0 & 0 & 2 & 2 & 2 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 3 & 2 & 2 & 1 & 0 & 2 & 2 & 1 \\ 0 & 3 & 2 & 3 & 1 & 0 & 3 & 2 & 2 & 2 & 0 & 2 & 2 & 1 \\ 2 & 1 & 2 & 0 & 3 & 2 & 2 & 2 & 1 & 1 & 0 & 3 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 3 & 0 & 0 & 3 & 3 & 3 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 0 & 3 & 3 & 1 & 0 & 3 & 3 & 1 \\ 0 & 3 & 2 & 3 & 1 & 0 & 1 & 3 & 0 & 2 & 3 & 3 & 3 & 1 \\ 1 & 0 & 3 & 0 & 1 & 0 & 0 & 3 & 3 & 3 & 0 & 3 & 3 & 1 \\ 3 & 1 & 0 & 2 & 2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ 1 & 0 & 1 & 0 & 2 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 2 \\ 1 & 3 & 2 & 0 & 2 & 0 & 1 & 0 & 0 & 2 & 0 & 0 & 0 & 2 \\ 1 & 0 & 3 & 0 & 2 & 0 & 1 & 0 & 0 & 3 & 0 & 0 & 0 & 2 \\ 3 & 1 & 0 & 2 & 2 & 0 & 1 & 1 & 0 & 0 & 2 & 1 & 2 & 2 \\ 1 & 0 & 1 & 0 & 2 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 2 & 2 \\ 1 & 3 & 2 & 0 & 2 & 0 & 3 & 1 & 2 & 2 & 0 & 1 & 2 & 2 \\ 1 & 0 & 3 & 0 & 2 & 0 & 0 & 1 & 3 & 3 & 0 & 1 & 2 & 2 \\ 3 & 1 & 0 & 2 & 2 & 0 & 1 & 2 & 0 & 0 & 2 & 2 & 2 & 2 \\ 1 & 0 & 1 & 0 & 2 & 0 & 3 & 2 & 2 & 1 & 0 & 2 & 2 & 2 \\ 1 & 3 & 2 & 0 & 2 & 0 & 3 & 2 & 2 & 2 & 0 & 2 & 2 & 2 \\ 1 & 0 & 3 & 0 & 2 & 2 & 1 & 2 & 0 & 3 & 2 & 2 & 2 & 2 \\ 3 & 1 & 0 & 2 & 2 & 0 & 1 & 3 & 0 & 0 & 0 & 3 & 0 & 2 \\ 1 & 0 & 1 & 0 & 2 & 0 & 0 & 3 & 3 & 1 & 0 & 3 & 0 & 2 \\ 1 & 3 & 2 & 0 & 2 & 0 & 1 & 3 & 0 & 2 & 0 & 3 & 0 & 2 \\ 1 & 0 & 3 & 0 & 2 & 0 & 0 & 3 & 3 & 3 & 0 & 3 & 0 & 2 \\ 0 & 1 & 0 & 3 & 3 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \\ 1 & 0 & 1 & 0 & 3 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 3 \\ 0 & 3 & 2 & 3 & 3 & 0 & 1 & 0 & 0 & 2 & 0 & 0 & 0 & 3 \\ 1 & 0 & 3 & 0 & 3 & 0 & 1 & 0 & 0 & 3 & 0 & 0 & 0 & 3 \\ 0 & 1 & 0 & 3 & 3 & 0 & 1 & 1 & 0 & 0 & 3 & 1 & 3 & 3 \\ 1 & 0 & 1 & 0 & 3 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 3 & 3 \\ 0 & 3 & 2 & 3 & 3 & 0 & 3 & 1 & 2 & 2 & 3 & 1 & 3 & 3 \\ 1 & 0 & 3 & 0 & 3 & 0 & 0 & 1 & 3 & 3 & 0 & 1 & 3 & 3 \\ 0 & 1 & 0 & 3 & 3 & 0 & 1 & 2 & 0 & 0 & 0 & 2 & 0 & 3 \\ 1 & 0 & 1 & 0 & 3 & 0 & 3 & 2 & 2 & 1 & 0 & 2 & 0 & 3 \\ 0 & 3 & 2 & 3 & 3 & 0 & 3 & 2 & 2 & 2 & 0 & 2 & 0 & 3 \\ 1 & 0 & 3 & 0 & 3 & 0 & 1 & 2 & 0 & 3 & 0 & 2 & 0 & 3 \\ 0 & 1 & 0 & 3 & 3 & 0 & 1 & 3 & 0 & 0 & 3 & 3 & 3 & 3 \\ 1 & 0 & 1 & 0 & 3 & 0 & 0 & 3 & 3 & 1 & 0 & 3 & 3 & 3 \\ 0 & 3 & 2 & 3 & 3 & 0 & 1 & 3 & 0 & 2 & 3 & 3 & 3 & 3 \\ 1 & 0 & 3 & 0 & 3 & 0 & 0 & 3 & 3 & 3 & 0 & 3 & 3 & 3 \\ \hline \end{array}$

Hence according to the definition of $\mathscr C_5$, $(A4)$ is not a tautology.

Therefore $(A4)$ is independent from $(A1)$, $(A2)$, $(A3)$.