Groups are numbers (2). Pattern matching

As in divination, pattern matching. Continues from Groups are numbers (1).

We start from elementary variables, then we define number terms by two operations: substraction and multiplication.

accept_0_0

  • Variables are terms.
  • Substraction (the first line): a is a variable and b is a term, then a-b is a term.
  • Multiplication (2nd line): a, b are terms, then ab is a term.

 

By pattern matching we can prove for example this:

accept_4

[update: figure replaced, the initial one was wrong by pattern matching only. The difference is that in this correct figure appears “(a-b)d” instead of the wrong “d(a-b)”]

What does it mean? These are just binary trees. Well let’s take a typing convention

accept_0_1

where e, x, … are elements of a vector space and the variables are invertible scalars. Moreover take e = 0 for simplicity.

Then the previous pattern matching dream says that

(1-(a-b))c x + (a-b)(c-d) x = (c - (a-b)d)x

which is true, but from all the irrelevant reasons (vector space, associativity and commutativity  of addition, distributivity, etc):

(c- ac + bc + ac -ad - bc + bd) x = (c - ad + bd) x = (c - (a-b)d)x 

What about this one, which is also by pattern matching:

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With the previous typing conventions it reads:

(c-(a-b))x = (1-b)(c-a)x + b(1- a^{-1})(c-a) x + (bc a^{-1})x

which is true because the right hand side is:

((1-b)(c-a) + b(1- a^{-1})(c-a) + bc a^{-1} )x =

= (c-a-bc+ab+bc-ab-bca^{-1} +b+bc a^{-1}) x = (c-a+b) x = (c-(a-b))x

Which is funny because it does not make any sense.

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