metamathematics - Bourbaki's epsilon-calculus notation

Matthias' polemics are funny at points but also misleading in several respects:

1. ZFC also has enormous length and depth of deductions for trivial material. According to Norman Megill's metamath page, "complete proof of 2 + 2 = 4 involves 2,452 subtheorems including the 150 [depth of the proof tree] above. ... These have a total of 25,933 steps — this is how many steps you would have to examine if you wanted to verify the proof by hand in complete detail all the way back to the axioms." Megill's system is based on a formalism for substitutions so there may be an enormous savings here compared to the way in which Matthias performs the counts (i.e., the full expanded size in symbols) for Bourbaki's system. If I correctly recall other information from Megill about the proof length he estimated for various results in ZFC, the number of symbols required can be orders of magnitude larger and this is what should be compared to Matthias' numbers.




2. The proof sizes are enormously implementation dependent. Bourbaki proof length could be a matter of inessential design decisions. Matthias claims at the end of the article that there is a problem using Hilbert epsilon-notation for incomplete or undecidable systems, but he gives no indication that this or any other problem is insurmountable in the Bourbaki approach.




3. Indeed, Matthias himself appears to have surmounted the problem in his other papers, by expressing Bourbaki set theory as a subsystem of ZFC. So either he has demonstrated that some reasonably powerful subsystems of ZFC have proofs and definitions that get radically shorter upon adding Replacement, or that the enormous "term" he attributes to the Theorie des Ensembles shrinks to a more ZFC-like size when implemented in a different framework.

EDIT. A search for Norman Megill's calculations of proof lengths in ZFC found the following:

"even trivial proofs require an
astonishing number of steps directly from axioms. Existence of the
empty set can be proved with 11,225,997 steps and transfinite recursion
can be proved with 11,777,866,897,976 steps."

and

"The proofs exist only in principle, of course, but their
lengths were backcomputed from what would result from more traditional
proofs were they fully expanded. ..... In the current version of my proof
database which has been reorganized somewhat, the numbers are:

empty set = 6,175,677 steps

transf. rec. = 24,326,750,185,446 steps"

That's only the number of steps. The number of symbols would be much, much higher.