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The obstruction to acquiring a binary phase singleton

Abstract

Binary initial phase singletons require two complete blocks and paid arrival.

Theorem 1.1 (Every correct endpoint tree has horizon at least max(2,ceil(j/m))).

Proof. Machine-checked in Lean as D5/S3/ObserverMemory/Algorithms/KBonacciAcquisition/CoprimeSingletonLower.singleton_tree_obstruction (✓ std3). ∎

Source. Repository-derived.

Commentary.

For arbitrary Y, distinct labels A and B, value v in ZMod(2), either alphabet, k>=2, 2<=m<k and 1<=j<=k, every tree that labels all initial records (v,-i,s) with i<=k and s=max(2,ceil(j/m)). No coprimality assumption or separate survival premise is needed.

The marker is the indicator of zero phase with values in ZMod(2). Its adjacent sum is the literal coefficient; at phase -i+t for i,t<=k it is the indicator of t=i. Rejection is absorbing for every Boolean word. The casts in the formula take values in ZMod(k+1), and all marker arithmetic takes values in ZMod(2).

The literal transitions are those of D5/S3/ObserverMemory/Algorithms/KBonacciAcquisition/LiteralModel and the trees have the endpoint semantics of D5/S3/ObserverMemory/Algorithms/KBonacciAcquisition/EndpointCells. A first bit equal to one would reject every record with tail k-1, so a correct tree starts with zero. Before arrival at j, the zero and target phases have identical observations and archives, forcing the arrival bound. A one-block singleton would have total increment one, whereas adjacent marker cancellation makes that total zero. This forces two blocks.

References