LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Counting bounded-size Boolean circuits

Statement

For $n, s \geq 1$ , the number of Boolean functions on $n$ inputs computed by circuits of size at most $s$ over the fixed page basis is at most $2^{O (s \log (n + s))} .$

Facts & Assumptions

Given: integers $n, s \geq 1$ .

[L1]

The circuit basis is fixed and has arity at most two, by Boolean circuits: basis, fan-in, size, and depth.

[L2]

Independent finite choices multiply, by The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ .

Proof

technique · direct

1.1

A topologically numbered gate has at most a fixed number $b$ of type choices and at most two input-wire indices among at most $n + s$ available wires. Thus each gate record has at most $b (n + s)^{2}$ possibilities; choosing an output wire adds at most $n + s$ possibilities.

L1given

2.1

By [L2], circuits with exactly $r \leq s$ gates therefore number at most $(n + s) [b (n + s)^{2}]^{r}$ . Summing over $0 \leq r \leq s$ gives at most $(s + 1) (n + s) [b (n + s)^{2}]^{s} = 2^{O (s \log (n + s))}$ because $s \geq 1$ .

L2step 1.1algebra

3.1

Distinct encodings may compute the same function, so this circuit count is also an upper bound on the number of represented Boolean functions.

step 2.1∎

Depends on

Used by

Shannon counting gives an efficiently explicit circuit-hard function False statement
Almost all Boolean functions require exponential circuit size Theorem

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Arora and Barak, Computational Complexity: A Modern Approach (standard reference, not scraped)
Lance Fortnow, Counting Complexity (standard reference, not scraped)