Denote with $\mathscr H_2 - (\text A 4)$ the proof system resulting from $\mathscr H_2$ by removing axiom $(\text A 4)$. Consider $\mathscr C_5$, Instance 5 of constructed semantics. We will prove that: * $\mathscr H_2 - (\text A 4)$ is sound for $\mathscr C_5$; * Axiom $(\text A 4)$ is not a tautology in $\mathscr C_5$ which leads to the conclusion that $(\text A 4)$ is not a theorem of $\mathscr H_2 - (\text A 4)$. === Soundness of $\mathscr H_2 - (\text A 4)$ for $\mathscr C_5$ === Starting with the axioms: {{begin-axiom}} {{axiom | n = \text A 1         | lc= Rule of Idempotence         | m = \paren {p \lor p} \implies p         | rc= Proof of Tautology }} {{axiom | n = \text A 2         | lc= Rule of Addition         | m = q \implies \paren {p \lor p}         | rc= Proof of Tautology }} {{axiom | n = \text A 3         | lc= Rule of Commutation         | m = \paren {p \lor q} \implies \paren {q \lor p}         | rc= Proof of Tautology }} {{end-axiom}} Next it needs to be shown that the rules of inference of $\mathscr H_2$ preserve $\mathscr C_5$-tautologies. ==== Rule $\text {RST} 1$: Rule of Uniform Substitution ==== By definition, any WFF is assigned a value from $\set {0, 1, 2, 3}$. Thus, in applying Rule $\text {RST} 1$, 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 $\text {RST} 1$ preserves $\mathscr C_5$-tautologies. ==== Rule $\text {RST} 2$: Rule of Substitution by Definition ==== Because the definition of $\mathscr C_5$ was given in terms of Rule $\text {RST} 2$, it cannot affect any of its results. ==== Rule $\text {RST} 3$: Rule of Detachment ==== Suppose $\mathbf A$ and $\mathbf A \implies \mathbf B$ both take value $0$. Then using Rule $\text {RST}  2$, 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 $\text {RST} 3$ also produces only WFFs of value $0$. ==== Rule $\text {RST} 4$: Rule of Adjunction ==== Suppose $\mathbf A$ and $\mathbf B$ take value $0$. Then using the definitional abbreviations: :$\mathbf A \land \mathbf B =_{\text {def} } \neg \paren {\neg \mathbf A \lor \neg \mathbf B}$ We compute: {{begin-eqn}} {{eqn | l = \mathbf A \land \mathbf B       | r = 0 \land 0 }} {{eqn | r = \neg \paren {\neg 0 \lor \neg 0}       | c = Rule $\text {RST} 2 \, (1)$ }} {{eqn | r = \neg \paren {1 \lor 1} }} {{eqn | r = \neg 1 }} {{eqn | r = 0 }} {{end-eqn}} proving that Rule $\text {RST} 4$ also produces only $0$s from $0$s. Hence $\mathscr H_2 - (\text A 4)$ is sound for $\mathscr C_5$. === $(\text A 4)$ is not a $\mathscr C_5$-tautology === Recall axiom $(\text A 4)$, the Factor Principle: :$\paren {p \lor q} \implies \paren {q \lor p}$ Under $\mathscr C_5$, we can use definitional abbreviations to arrive at: :$\neg \paren {\neg p \lor q} \lor \paren {\neg \paren {r \lor p} \lor \paren {r \lor q} }$ 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 $\mathscr C_5$, $(\text A 4)$ is not a tautology. Therefore $(\text A 4)$ is independent from $(\text A 1)$, $(\text A 2)$, $(\text A 3)$. {{qed}}