Pascal's Triangle is the Fold Lattice

The diagonal sums of Pascal's triangle are the Fibonacci numbers.

Row 0:  1                           sum of diagonal 0: 1 = F(2)
Row 1:  1  1                        sum of diagonal 1: 1 = F(3)? No...
Row 2:  1  2  1
Row 3:  1  3  3  1
Row 4:  1  4  6  4  1

The precise statement: F(n+1) = Σ C(n-k, k) for k = 0 to ⌊n/2⌋. The shallow diagonals of Pascal's triangle -- not the rows, not the columns, but the diagonals running from upper-right to lower-left -- sum to consecutive Fibonacci numbers. Proved for n = 0 through 8 in FibonacciDeep2.lean.

Why does this matter? Because C(n, k) -- "n choose k" -- is the number of ways to fold k items out of n. Pascal's triangle is the complete catalog of all possible folds. Every entry counts a different way to commit to a subset.

Fibonacci emerges from the diagonals of this catalog. Not from the rows (which count all subsets of a fixed size). Not from the columns (which count how one element participates across set sizes). From the diagonals -- which count the folds where each step skips one possibility. The non-consecutive constraint. The Zeckendorf condition. The same constraint that governs the complement distribution.

Fibonacci doesn't just model folds. Fibonacci formalizes the structure of all possible folds, counted along the dimension where consecutive choices are forbidden.

Formal proofs: fibonacci_pascals_triangle.test.gg + FibonacciDeep2.lean S5 (28 theorems) Next section: The Golden Ratio Identities