Denote with $\mathscr H_2 - (A3)$ the proof system resulting from $\mathscr H_2$ by removing axiom $(A3)$. Consider $\mathscr C_4$, Instance 4 of constructed semantics. We will prove that: * $\mathscr H_2 - (A3)$ is sound for $\mathscr C_4$; * Axiom $(A3)$ is not a tautology in $\mathscr C_4$ which leads to the conclusion that $(A3)$ is not a theorem of $\mathscr H_2 - (A3)$. === Soundness of $\mathscr H_2 - (A3)$ for $\mathscr C_4$ === Starting with the axioms: {{begin-axiom}} {{axiom|n = A1        |lc = Rule of Idempotence        |m = (p \lor p) \implies p        |rc = Proof of Tautology }} {{axiom|n = A2        |lc = Rule of Addition        |m = q \implies (p \lor q)        |rc = Proof of Tautology }} {{axiom|n = A4        |lc = Factor Principle        |m = (q \implies r) \implies \left({ (p \lor q) \implies (p \lor r)}\right)        |rc = Proof of Tautology }} {{end-axiom}} Next it needs to be shown that the rules of inference of $\mathscr H_2$ preserve $\mathscr C_4$-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 $\mathscr C_4$-tautologies. ==== Rule $RST \, 2$: Rule of Substitution by Definition ==== Because the definition of $\mathscr C_4$ 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: {{begin-eqn}} {{eqn|l = \mathbf A \land \mathbf B      |r = 0 \land 0 }} {{eqn|r = \neg ( \neg 0 \lor \neg 0 )      |c = Rule $RST \, 2 \, (1)$ }} {{eqn|r = \neg ( 1 \lor 1 ) }} {{eqn|r = \neg 1 }} {{eqn|r = 0 }} {{end-eqn}} proving that Rule $RST \, 4$ also produces only $0$s from $0$s. Hence $\mathscr H_2 - (A3)$ is sound for $\mathscr C_4$. === $(A3)$ is not a $\mathscr C_4$-tautology === Recall axiom $(A3)$, the Rule of Commutation: :$(p \lor q) \implies (q \lor p)$ Under $\mathscr C_4$, we apply a single definitional abbreviation and have the following: ::$\begin{array}{|cccc|c|ccc|} \hline \neg & (p & \lor & q) & \lor & (q & \lor & p) \\ \hline 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 2 & 0 & 2 & 0 & 0 \\ 1 & 0 & 0 & 3 & 0 & 3 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 2 & 2 & 0 & 2 & 2 & 1 \\ 2 & 1 & 3 & 3 & 0 & 3 & 3 & 1 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 2 & 2 & 1 & 0 & 1 & 2 & 2 \\ 0 & 2 & 2 & 2 & 0 & 2 & 2 & 2 \\ 1 & 2 & 0 & 3 & 3 & 3 & 3 & 2 \\ 1 & 3 & 0 & 0 & 0 & 0 & 0 & 3 \\ 2 & 3 & 3 & 1 & 0 & 1 & 3 & 3 \\ 2 & 3 & 3 & 2 & 0 & 2 & 0 & 3 \\ 2 & 3 & 3 & 3 & 0 & 3 & 3 & 3 \\ \hline \end{array}$ Hence according to the definition of $\mathscr C_4$, $(A3)$ is not a tautology. Therefore $(A3)$ is independent from $(A1)$, $(A2)$, $(A4)$. {{qed}}