I've begun studying mathematical logic with Hodel's 'An Introduction to Mathematical Logic', and have just learnt about formal systems and their components. In the textbook, Hodel gives as an example of a formal system the system 'ADD', whose theorems correspond to true statements about addition; it is defined as follows:
- Symbols: $| \,,+,=$
- Formulae: statements of the form $x+y=z$ where $x,y$, and $z$ are expressions in $|$
- Axiom: $|+|=||$
- Rules of Inference: $\frac{x+y=z}{x|+y=z|}$ (R1) and $\frac{x+y=z}{y+x=z}$ (R2)
As an exercise (exercise 3 of section 1.3 for those with the book), Hodel then asks the reader to create a formal system 'MULT', which has theorems corresponding to our standard notions of multiplication; he also asks the reader to prove within this system that $|| \times ||| = ||||||$.
I believe I have a solution to this, but I would be grateful if someone wouldn't mind checking it for me - I'm new to this area of maths so want to make sure my fundamentals are strong. My system is defined as follows:
- Symbols: $|\, , \times, + , =$
- Formulae: statements of the form $x\times y = z$ or $x+y=z$ where $x, y,$ and $z$ are expressions in $|$
- Axioms: $|+|=||$ (A1) and $| \times x = x$ (A2)
- Rules of inference: $\frac{x+y=z}{x|+y=z|}$ (R1), $\frac{x+y=z}{y+x=z}$ (R2), $\frac{x\times y=z}{x|\times y=z+y}$ (R3), and $\frac{x\times y=z}{y\times x=z}$ (R4)
and my proof that $\vdash_{\text{MULT}}|| \times ||| = ||||||$:
- $|\times||=||$ (A2)
- $||\times || = ||+||$ (R3)
- $|+|=||$ (A1)
- $||+| = |||$ (R1)
- $|+|| = |||$ (R2)
- $||+|| = |||| = || \times || $(R1 and step 2)
- $||| \times || = ||||+||$ (R3)
- $|| + || = ||||$ (step 6 and previous)
- $||||+|| = |||||| = ||| \times ||$ (R1 twice and step 7)
- $ || \times ||| = ||||||$ (R4) $\square$
What makes me doubt whether this is the simplest solution is that I have included all of ADD within my new system, however I suppose this does reflect how multiplication is defined in terms of iterated addition.
Improvements, corrections, and other solutions welcomed and greatly appreciated :)
Here's a solution: Reformulate A2 as: I x I = I. Then, the proof goes as follows:
I x I = I (A2)
II x I = II (R3)
III x I = III (R3)
I x III = III (R4)
II x III = IIIIII (R3)