Hilbert Proof System Instance 2 Independence Results/Independence of A4

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Theorem

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

Then:


Axiom $(\text A 4)$ is independent from $(\text A 1)$, $(\text A 2)$, $(\text A 3)$.


Proof

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:

\((\text A 1)\)   $:$   Rule of Idempotence    \(\ds \paren {p \lor p} \implies p \)      Proof of Tautology
\((\text A 2)\)   $:$   Rule of Addition    \(\ds q \implies \paren {p \lor p} \)      Proof of Tautology
\((\text A 3)\)   $:$   Rule of Commutation    \(\ds \paren {p \lor q} \implies \paren {q \lor p} \)      Proof of Tautology


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:

\(\ds \mathbf A \land \mathbf B\) \(=\) \(\ds 0 \land 0\)
\(\ds \) \(=\) \(\ds \neg \paren {\neg 0 \lor \neg 0}\) Rule $\text {RST} 2 \, (1)$
\(\ds \) \(=\) \(\ds \neg \paren {1 \lor 1}\)
\(\ds \) \(=\) \(\ds \neg 1\)
\(\ds \) \(=\) \(\ds 0\)

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 $(\text A 1)$, $(\text A 2)$, $(\text A 3)$.

$\blacksquare$


Sources

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