- #141
atyy
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suremarc said:There is no distinction between countable and uncountable in PRA itself. No uncountable ordinal exists in PRA (its proof theoretic ordinal is ##\omega^{\omega}##).
Nor does recursion involve quantification over the natural numbers, which is a condition of the incompleteness theorems.
I replied to this in post #140, and as I said my main point in bringing up PRA had nothing to do with the incompleteness theorem. However, it does seem that PRA is also incomplete http://www.personal.psu.edu/t20/notes/logic.pdf (Theorem 6.6.4, p122).