imports/sets/chapter_1/1.2.md read against ens-i-iv
4 printed pages, 2 sentences of the pages not in the import, 2 sentences of the import on no page.
share audit has already held this import against the printed contents and the printed labels, and this sheet does not repeat that. What is here is the prose: the sentences of each printed page that the import does not carry, and the sentences of the import that no page of the section carries. A passage on this sheet is a place to look and not a fault. The two sides are two transcriptions of one piece of type and they disagree about markup, about headings and about running heads without either being wrong about the book.
pdf page 31, 97% of its prose in the import
Nothing to look at. 4 sentences under eight words, too short to make a run and left out.
pdf page 32, 94% of its prose in the import
Nothing to look at. 5 sentences under eight words, too short to make a run and left out.
pdf page 33, 100% of its prose in the import
Nothing to look at. 6 sentences under eight words, too short to make a run and left out.
pdf page 34, 96% of its prose in the import
- 47% found. \*For example, in the theory of groups $\mathscr{T}$, the explicit axioms contain two constants G and $\mu$ (the group and the law of composition).
- 36% found. If we substitute these terms for G and $\mu$ respectively in the explicit axioms of $\mathscr{T}$, we obtain theorems in $\mathscr{T}'$.
3 sentences under eight words, too short to make a run and left out.
In the import and on no page of the section
This is the half share audit cannot see. A transcription made by a model reading a photograph does not leave a gap where it could not read a line, it writes the sentence the book would most likely have there, and a fluent invention passes every check this project owns. Read each of these against the printed section before the import is promoted.
- 40% found. If we substitute these terms for $G$ and $\mu$ respectively in the explicit axioms of $\mathcal T$, we obtain theorems in $\mathcal T'$.
- 33% found. For if $A$ and "not $A$" are theorems in $\mathcal T$, then $$ (T_1\mid a_1)\cdots(T_h\mid a_h)A $$ and $$ \operatorname{not}(T_1\mid a_1)\cdots(T_h\mid a_h)A $$ are theorems in $\mathcal T'$.
19 sentences under eight words, too short to make a run and left out.