pvsNP_one_statement
The declaration asserts that the BIT substrate has per-cycle bandwidth exactly 360, that total bandwidth after t cycles is strictly less than the 2^n workload whenever 360 t < 2^n, and that workload doubles with each added input bit. Complexity theorists studying recognition substrates would cite it for the exponential certification lower bound on any BIT-compatible system. The proof is a one-line term that packages the bandwidth equality, the cycles lower bound lemma, and the workload doubling lemma.
claimLet $B=360$ be the per-cycle BIT bandwidth. Then $B=360$, and for all natural numbers $n,t$, if $B t < 2^n$ then the total bandwidth in $t$ cycles is strictly less than the minimum number of distinguishable comparisons $2^n$ required to certify an NP-search witness of size $n$, and the workload satisfies $2^{n+1}=2·2^n$.
background
In the Recognition Science setting the recognition operator on a finite-state substrate performs at most $B·t$ useful comparisons in $t$ cycles, where $B$ is the BIT bandwidth. The module defines bitBandwidthPerCycle as $8·45=360$ and bandwidthBudget $t$ as $B·t$. The npWorkload $n$ is defined as $2^n$, the canonical search-space size for an NP problem of input length $n$. The upstream cycles_lower_bound theorem states that $360 t < 2^n$ implies bandwidthBudget $t < npWorkload n$, while npWorkload_succ shows that adding one bit exactly doubles the workload.
proof idea
The proof is a one-line term that directly supplies the three conjuncts of the conjunction: the equality bitBandwidthPerCycle_eq (proved by norm_num), the cycles_lower_bound theorem (which applies insufficient_budget_no_certify after rewriting bandwidth), and the npWorkload_succ theorem (proved by unfolding and using pow_succ with ring).
why it matters in Recognition Science
This packages the key structural lower bound for NP certification time on BIT substrates inside the Recognition Science framework, completing the derivation in the Complexity.PvsNPFromBIT module (Track F8). It supplies the exponential-in-$n$ cost that follows from the recognition operator and the bounded per-cycle bandwidth; the result is substrate-dependent and does not claim a classical Turing separation of P and NP. It touches the open question whether any physical substrate can exceed the BIT bandwidth limit without violating the recognition axioms.
scope and limits
- Does not separate P from NP in the classical Turing model.
- Does not apply to substrates permitting oracles or hypercomputation.
- Does not supply an upper bound on certification time.
- Does not incorporate quantum or non-classical computational effects.
- Does not address average-case or smoothed complexity.
formal statement (Lean)
143theorem pvsNP_one_statement :
144 bitBandwidthPerCycle = 360 ∧
145 (∀ {n t : ℕ}, 360 * t < 2 ^ n → bandwidthBudget t < npWorkload n) ∧
146 (∀ n, npWorkload (n + 1) = 2 * npWorkload n) :=
proof body
Term-mode proof.
147 ⟨bitBandwidthPerCycle_eq, @cycles_lower_bound, npWorkload_succ⟩
148
149end
150
151end PvsNPFromBIT
152end Complexity
153end IndisputableMonolith