Denote with $\mathscr H_2 - (A2)$ the proof system resulting from $\mathscr H_2$ by removing axiom $(A2)$. Consider $\mathscr C_3$, Instance 3 of constructed semantics. We will prove that: * $\mathscr H_2 - (A2)$ is sound for $\mathscr C_3$; * Axiom $(A2)$ is not a tautology in $\mathscr C_3$ which leads to the conclusion that $(A2)$ is not a theorem of $\mathscr H_2 - (A2)$. === Soundness of $\mathscr H_2 - (A2)$ for $\mathscr C_3$ === 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 = A3        |lc = Rule of Commutation        |m = (p \lor q) \implies (q \lor p)        |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 $\mathscr C_3$-tautologies. ==== Rule $RST \, 1$: Rule of Uniform Substitution ==== By definition, any WFF is assigned a value $0$, $1$ or $2$. Thus, in applying Rule $RST \, 1$, we are introducing $0$, $1$ or $2$ in the position of a propositional variable. But all possibilities of assignments of $0$s, $1$s and $2$s to such propositional variables were shown not to affect the resulting values of the axioms. Hence Rule $RST \, 1$ preserves $\mathscr C_3$-tautologies. ==== Rule $RST \, 2$: Rule of Substitution by Definition ==== Because the definition of $\mathscr C_3$ 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 $0$ or $1$. Then using Rule $RST \, 2$, definition $(2)$, we get: :$\neg \mathbf A \lor \mathbf B$ taking value $0$ or $1$ by assumption. But $\neg \mathbf A$ takes value $1$ or $2$, by definition of $\neg$. So from the definition of $\lor$, we have the following options: $\begin{array}{|c|c|c|} \hline \mathbf A & \neg \mathbf A \lor \mathbf B & \\ \hline 0 & 0 & 2 \lor \mathbf B = 0 \implies \mathbf B = 0\\ 1 & 0 & 1 \lor \mathbf B = 0 \implies \mathbf B = 0\\ 0 & 1 & 2 \lor \mathbf B = 1 \implies \mathbf B = 1\\ 1 & 1 & 1 \lor \mathbf B = 1 \implies \mathbf B = 1\\ \hline \end{array}$ In each case, we see that the definition of $\lor$ necessitates that $\mathbf B$ takes value $0$ or $1$, respectively. Hence Rule $RST \, 3$ also produces only WFFs of value $0$ or $1$. ==== Rule $RST \, 4$: Rule of Adjunction ==== Suppose $\mathbf A$ and $\mathbf B$ take value $0$ or $1$. Then using the definitional abbreviations: :$\mathbf A \land \mathbf B =_{\text{def}} \neg ( \neg \mathbf A \lor \neg \mathbf B )$ We compute: :$\begin{array}{|c|ccccc|} \hline \neg & ( \neg & \mathbf A & \lor & \neg & \mathbf B ) \\ \hline 0 & 2 & 0 & 2 & 2 & 0 \\ 0 & 2 & 0 & 2 & 1 & 1 \\ 0 & 1 & 1 & 2 & 2 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$ proving that Rule $RST \, 4$ also produces only $0$s and $1$s from $0$s and $1$s. Hence $\mathscr H_2 - (A2)$ is sound for $\mathscr C_3$. === $(A2)$ is not a $\mathscr C_3$-tautology === Recall axiom $(A2)$, the Rule of Addition: :$q \implies (p \lor q)$ Under $\mathscr C_3$, we apply a single definitional abbreviation and have the following: :$\begin{array}{|cc|c|ccc|} \hline \neg & q & \lor & (p & \lor & q) \\ \hline 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & 0 & 0 & 2 \\ 2 & 0 & 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 2 & 0 & 1 & 2 & 2 \\ 2 & 0 & 0 & 2 & 0 & 0 \\ 1 & 1 & 2 & 2 & 2 & 1 \\ 0 & 2 & 0 & 2 & 2 & 2 \\ \hline \end{array}$ Hence according to the definition of $\mathscr C_3$, $(A1)$ is not a tautology. Therefore $(A2)$ is independent from $(A1)$, $(A3)$, $(A4)$. {{qed}}