joint_125_squared
plain-language theorem explainer
The declaration establishes that the product of two 125-element state spaces equals exactly 15625, realizing the 5^6 joint lattice for cognitive and oncology domains in the Recognition Science product hierarchy. Cross-domain modelers cite it to anchor the information-theoretic bound on combined patient state size. The proof reduces to a single decide tactic that evaluates the arithmetic equality by direct computation.
Claim. The product of two copies of the 125-element state space satisfies $125 times 125 = 15625$.
background
The module defines a lattice of product state spaces built from the base triple 5^3 = 125, drawn from cognitive (C1) and oncology (C3) domains. The hierarchy runs 5^2 = 25 for domain pairs, 5^3 = 125 for single triples, 5^4 = 625 for four-fold products, 5^5 = 3125 for full domain stacks, and 5^6 = 15625 for the joint cognitive-oncology state. This joint size is required to satisfy the RS bound 5^6 < 2^14 = 16384, keeping the combined state within 14 bits.
proof idea
The proof is a one-line wrapper that applies the decide tactic to verify the concrete numerical equality 125 * 125 = 15625 by exhaustive arithmetic reduction.
why it matters
This fills the 5^6 slot in the product recognition lattice, confirming the joint state size for cognitive times oncology spaces under the cross-domain hierarchy. It directly supports the module's structural claim that products of 5^3 triples remain bounded by the 14-bit information limit. No downstream theorems are listed, so the result serves as a concrete anchor for the larger-joints section rather than feeding a named parent theorem.
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