closed
plain-language theorem explainer
Ledger closure on a cube-face plaquette: four vertex recognition bits have even parity (XOR to false). Anyone counting admissible boundary sectors on the forced 8-tick D=3 lattice cites this as the sole admissibility filter before quotienting by the square stabilizer. The definition is a one-line Boolean: negate the XOR of the four vertex bits packed in a Fin 16 configuration.
Claim. For a boundary face configuration $c$ (one bit on each of the four vertices of a cube face), the configuration is closed when $b_0 \oplus b_1 \oplus b_2 \oplus b_3 = 0$ (even parity), where $b_i$ is the recognition bit at vertex $i$. Equivalently, the Boolean predicate returns true precisely when the four bits XOR to false.
background
PixelLocal works on the forced discrete substrate of Recognition Science holography: the D=3, eight-tick cell is the cube $2^3$ (8 vertices, 6 faces). A boundary recognition pixel is one square face. Configurations place one recognition bit on each of the four vertices and pack them into the low four bits of a value in $\mathrm{Fin},16$.
The module isolates two structures only. Ledger closure is the first: a closed recognition loop must post a balanced (zero-sum) ledger around the plaquette, so the four vertex bits have even parity. This condition comes from the recognition ledger, not from continuum geometry. The second structure (used downstream) is the face stabilizer $D_4$, the eight symmetries of the square.
Vertex-bit extraction is the same pattern used on full cells and on glued domino plaquettes: test the $i$-th bit of the packed integer. The glued-domino analogue requires even parity on both unit faces simultaneously; here the predicate is the single-face case.
proof idea
Pure definition, not a theorem. Unpack the four vertex bits of $c$ via the bit-test map, XOR them in cyclic order $0,1,2,3$, and return the Boolean negation so that true means even parity (XOR equals false). No lemmas, no tactics: a one-line Boolean expression on the packed $\mathrm{Fin},16$ configuration.
why it matters
This is the sole admissibility gate in the greenlit construction that forces the recognition-sector count on a boundary pixel. After filtering closed configurations and quotienting by the square stabilizer $D_4$, the orbit count is exactly 4, i.e. $2^{D-1}$ realized on the forced lattice. That integer is the substrate-native stand-in for the geometric coefficient written as 4 in $a_{\mathrm{pix}} = 4 \cdot H \cdot \ell_P^2$, separated from the already-proved entropy $H$ and from the open Planck-area scale.
Downstream, admissible-sector enumerations, face-stabilizer actions, and sector-count theorems in this module all route through the predicate. Related holography and algebra sites reuse the same even-parity language (cell face records, domino closure, $\mathbb{F}2$ weight counts). Framework landmarks: T7 eight-tick octave and T8 $D=3$ fix the cube; the count itself does not import Bekenstein–Hawking or continuum isotropy. Open gap flagged by the module: linking this sector count to the area coefficient in $a{\mathrm{pix}}$ remains a separate argument.
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