Hilbert Proof System Instance 2 Independence Results/Independence of A3

== Theorem ==
Let $\mathscr H_2$ be Instance 2 of the Hilbert proof systems.
Then:
Axiom $(A3)$ is independent from $(A1)$, $(A2)$, $(A4)$.
== Proof ==
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:
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:
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 $(A1)$, $(A2)$, $(A4)$.
== Sources ==
* : $\S 4.6$: Independence
Category:Hilbert Proof System Instance 2