Hilbert Proof System Instance 2 Independence Results/Independence of A2

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Theorem

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

Then:


Axiom $(A2)$ is independent from $(A1)$, $(A3)$, $(A4)$.


Proof

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:

\((A1)\)   $:$   Rule of Idempotence    \(\ds (p \lor p) \implies p \)      Proof of Tautology
\((A3)\)   $:$   Rule of Commutation    \(\ds (p \lor q) \implies (q \lor p) \)      Proof of Tautology
\((A4)\)   $:$   Factor Principle    \(\ds (q \implies r) \implies \left({ (p \lor q) \implies (p \lor r)}\right) \)      Proof of Tautology


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 $(A1)$, $(A3)$, $(A4)$.

$\blacksquare$


Sources

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