How to syntactically prove $\emptyset \vdash \top$?

$$\emptyset \vdash \top$$

I was trying to prove this using the Hilbert proof system and this is what I got:

(1)$$\top \equiv (\bot \equiv \bot)$$ Axiom:$$\top$$ vs. $$\bot$$

(2)$$(\top \equiv \top)\equiv (\top \equiv (\bot \equiv \bot))$$ by Leibniz

(3) $$(\top \equiv \top)$$ by equanimity on (2) and (1)

So this is where I am stuck. I confused on how I could use some axiom to isolate $$\top$$. Any help is much appreciated thank you! In my text-book if our assumptions are the $$\emptyset$$ then we can assume $$\top$$ is an absolute theorem but I am not sure how to go about proving this.

Axiom List: (https://i.sstatic.net/SjSH1.jpg)

Axioms of Boolean Logic $$\begin{array} \\ \text{ Associativity of } \equiv & ((A \equiv B) \equiv C) \equiv(A \equiv(B \equiv C)) \\ \text { Symmetry of } \equiv & (A \equiv B) \equiv(B \equiv A) \\ \text { Tvs. } \perp & T \equiv \perp \equiv \perp \\ \text { introduction of } \neg & \neg A \equiv A \equiv \perp \\ \text { Associativity of } \vee & (A \vee B) \vee C \equiv A\vee (B\vee C) \\ \text { Symmetry of } \vee & A \vee B \equiv B \vee A \\ \text { Idempotency of } \vee & A \vee A \equiv A \\ \text {Distributivity of } \vee \text{ over } \equiv & A \vee(B \equiv C) \equiv A \vee B \equiv A \vee C \\ \text { Excluded Middle } & A \vee \neg A \\ \text { Golden Rule } & A \wedge B \equiv A \equiv B \equiv A \vee B \\ \text { Implication } & A \rightarrow B \equiv A \vee B \equiv B \end{array}$$ Primary Rules of Inference $$\frac{A, A \equiv B}{B}\\~\\ \frac{A}{C[\mathbf{p}:=A] \equiv C[\mathbf{p}:=B]}$$

Book used: Mathematical Logic by George Tourlakis

• You should include a list of your axioms. There is no one standard system. Commented May 29, 2020 at 4:33
• I would like to help you, so I want to follow the axiom system from your textbook. What textbook are you studying? Commented May 29, 2020 at 5:07
• That system isn't at all "the Hilbert proof system". It looks like something some random guy came up with. Commented May 31, 2020 at 1:19

You need the following result: $$\vdash A \equiv A$$ (see page 47).

With it, you can complete your proof:

1) $$⊤ ≡ (⊥≡⊥)$$ --- Axiom

2) $$⊥≡⊥$$ --- result above

3) $$⊤$$ --- from 1) and 2) by (Eqn): $$\dfrac {A, A \equiv B}{B}$$,

the "equational-style" version of Modus Ponens.

• @VedarthV - you are welcome :-) Commented May 29, 2020 at 8:16