Note: I had to improvise a bit with notation, plus I reused a few symbols (letters for variables) sorry if it's long and complicated, I also hope I've done it all correctly

defined as follows

rule 1

i.e.

rule 2

i.e.

andotherwise

rule 3

I think I've come up with another way to express it

take

andand use the different levels of arithmetic

level 0

(successor function)level 1

(addition with B copies of 1)level 2

(multiplication with B copies of A)level 3

(exponentation with B copies of A)level 4

(tetration with B copies of A)and so on...

means level Y so means A to B level Y of arithmeticso based on this

with level 0 you do

Proof

first start with M=0

which is

then take M=1

according to rule 2according to rule 1 according to rule 3 according to rule 1so you get the sequence

etccompared to

etcso

and

according to rule 2according to the previous case according to the previous case

creating this sequence

compared to

which means

and

m=3

according to rule 2according to previous caseso with a first term of 5 and the operation multiply by 2 and then add 3

the formula will be of the form

multiply by 2 then add 3

and

so

and

so

so

now if L is used to mean a given level (where we know the formual works, and I've shown it works up to level 3) and LL means the next level

lets assume it works for a value of

first assumption according to rule 2and given the above rule

since

for any levelalso given that

substitute

according to first assumptionto generalise (is a second assumption, though it works for 0 and 1)

so

substituting

and

with (P+3) 2's)

with (P+4) 2's)

given that

and so on

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