Axioms of Hilbert Proof System Instance 1 for Predicate Logic are Tautologies

Theorem
Let $\mathscr H$ be Instance 1 of a Hilbert proof system for predicate logic.

Then the axioms of $\mathscr H$ are tautologies.

Axiom 1
This is precisely the statement of Propositional Tautology is Tautology in Predicate Logic.

By Well-Formed Formula is Tautology iff Universal Closure is Tautology, we prove the results for axioms 2 through 11 for the WFFs instead of their universal closures.

Let $\sigma$ be an assignment in a structure $\AA$.

Let $\map { \operatorname{val}_\AA } {\cdot} \sqbrk \sigma$ be the value mapping for $\sigma$ in $\AA$.

Axiom 2: $\mathbf A \implies \paren{ \forall x: \mathbf A }$
We have that $\map { \operatorname{val}_\AA } {\forall x: \mathbf A} \sqbrk \sigma = T$ :


 * $\forall a \in A: \map { \operatorname{val}_\AA } {\mathbf A} \sqbrk { \sigma + \paren{ a / x } } = T$

By assumption, $x$ does not occur freely in $\mathbf A$.

By Value of Formula under Assignment Determined by Free Variables, it follows that:


 * $\forall a \in A: \map { \operatorname{val}_\AA } {\mathbf A} \sqbrk { \sigma + \paren{ a / x } } = \map { \operatorname{val}_\AA } {\mathbf A} \sqbrk { \sigma }$

and hence:


 * $\map { \operatorname{val}_\AA } {\forall x: \mathbf A} \sqbrk \sigma = \map { \operatorname{val}_\AA } {\mathbf A} \sqbrk { \sigma }$

Therefore:

Since $\AA$ and $\sigma$ are arbitrary, it follows that $\mathbf A \implies \paren{ \forall x: \mathbf A }$ is a tautology.

Axiom 3: $\paren{ \forall x: \mathbf A \implies \mathbf B } \implies \paren{ \paren{ \forall x: \mathbf A } \implies \paren{ \forall x: \mathbf B } }$
that $\paren{ \forall x: \mathbf A \implies \mathbf B } \implies \paren{ \paren{ \forall x: \mathbf A } \implies \paren{ \forall x: \mathbf B } }$ is not a tautology.

Then there exist $\AA, \sigma$ such that:


 * $\map { \operatorname{val}_\AA } { \paren{ \forall x: \mathbf A \implies \mathbf B } \implies \paren{ \paren{ \forall x: \mathbf A } \implies \paren{ \forall x: \mathbf B } } } \sqbrk \sigma = F$

By definition of $\operatorname{val}_\AA$, this means:


 * $\map { f^\to } { \map { \operatorname{val}_\AA } {\forall x: \mathbf A \implies \mathbf B} \sqbrk \sigma, \map { \operatorname{val}_\AA } {\paren{ \forall x: \mathbf A } \implies \paren{ \forall x: \mathbf B } } \sqbrk \sigma } = F$

By definition of the truth function $f^\to$, this happens :


 * $\map { \operatorname{val}_\AA } {\forall x: \mathbf A \implies \mathbf B} \sqbrk \sigma = T$
 * $\map { \operatorname{val}_\AA } {\paren{ \forall x: \mathbf A } \implies \paren{ \forall x: \mathbf B } } \sqbrk \sigma = F$

and, applying the same argument on the second statement, we get:


 * $\map { \operatorname{val}_\AA } {\forall x: \mathbf A \implies \mathbf B} \sqbrk \sigma = T$
 * $\map { \operatorname{val}_\AA } {\forall x: \mathbf A } \sqbrk \sigma = T$
 * $\map { \operatorname{val}_\AA } {\forall x: \mathbf B } \sqbrk \sigma = F$

By definition of $\operatorname{val}_\AA$, the last of these statements means that for some $a \in A$:


 * $\map { \operatorname{val}_\AA } {\mathbf B } \sqbrk { \sigma + \paren{x / a} } = F$

where $\sigma + \paren{x / a}$ is the extension of $\sigma$ by mapping $x$ to $a$.

Similarly, from the first two statements it follows that:


 * $\map { \operatorname{val}_\AA } {\mathbf A } \sqbrk { \sigma + \paren{x / a} } = T$
 * $\map { \operatorname{val}_\AA } {\mathbf A \implies \mathbf B} \sqbrk { \sigma + \paren{ x / a} } = T$

However, by definition of $\operatorname{val}_\AA$:

Thus, by Proof by Contradiction, $\paren{ \forall x: \mathbf A \implies \mathbf B } \implies \paren{ \paren{ \forall x: \mathbf A } \implies \paren{ \forall x: \mathbf B } }$ is a tautology.