cycle_is_chiral
plain-language theorem explainer
The canonical 3-bit Gray-code cycle on the cube Q₃ is chiral: its per-axis bit-flip counts are not all equal. Anyone deriving CP violation, baryon asymmetry, or CKM structure from the eight-tick walk cites this. The proof is a short computational contradiction: bit 0 flips four times and bit 1 twice, so the counts cannot be S₃-invariant.
Claim. The Gray-code flip-count function on $Q_3$ is chiral: it is not the case that $\mathrm{count}(i)=\mathrm{count}(j)$ for all axes $i,j\in\{0,1,2\}$. Equivalently, the directed eight-tick walk does not treat the three cube axes symmetrically under permutation.
background
Recognition Science runs an eight-tick recognition operator along a directed Gray-code path on the 3-cube $Q_3$, namely the walk $[0,1,3,2,6,7,5,4]$. Each step flips exactly one bit; the flip sequence is $[0,1,0,2,0,1,0,2]$. The function $\mathrm{bitFlipCount}(b)$ counts how often axis $b$ is flipped over one full period.
A flip-count triple is called chiral when it fails to be constant on all three axes (equivalently, when it is not invariant under every permutation in $S_3$). The Gray-code counts are the function $\mathrm{grayFlipCounts}:=\mathrm{bitFlipCount}$. Module context: $J$-cost symmetry $J(x)=J(1/x)$ encodes CPT, while directed-cycle chirality is the geometric source of CP violation. Face-pairs of $Q_3$ correspond to the three particle generations, so unequal flip counts mean unequal generation couplings.
Upstream, the flip counts themselves are pure list-counting over the fixed Gray path; the chirality predicate is the negation of universal equality of those three naturals.
proof idea
Assume for contradiction that all three flip counts agree. Specialize to axes $0$ and $1$. Evaluate $\mathrm{bitFlipCount},0=4$ and $\mathrm{bitFlipCount},1=2$ by native_decide on the fixed Gray flip list. Unfold $\mathrm{grayFlipCounts}$ and rewrite; norm_num yields $4=2$, contradiction. Hence the universal-equality hypothesis is false, which is exactly chirality.
No external lemmas beyond the in-module count definitions are required; the argument is a finite computation on an 8-step path.
why it matters
This is the foundational chirality fact for CP violation in the Recognition framework. Downstream it is re-exported as cp_broken_by_chirality, paired with $J$-cost symmetry in cpt_ok_cp_broken (CPT preserved, CP broken), and bundled into the master chiralityCert.
Cosmology cites it inside derivation_chain_complete: three generations from $D=3$, chirality from this Gray path, positive Jarlskog invariant, and a positive baryon asymmetry. The Standard Model side uses the same $[4,2,2]$ asymmetry in ckmStructureCert for flavor-mixing structure.
Framework landmarks: T7 eight-tick octave (period $2^3$) supplies the walk length; T8 forces $D=3$, hence three axes and three generations. The $4:2:2$ split breaks residual $S_3$ axis symmetry while leaving $J(x)=J(1/x)$ intact, so CPT and CP-breaking coexist.
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