A universal code for three natural-number tracks

Goal and scope

N0 times {0,1,2} injects into N0 by c(n,j)=3n+j and is at most countable.

Finite-set certificates and a universal integer affine code; no uncountability conclusion.

Givens and symbols

• calc

• G1 — N0={n in Z:n≥0}; X=N0 times {0,1,2}; c(n,j)=3n+j.

• N_0 — natural numbers including zero
• c — integer affine code
• j — three-track index

Rules and sources

• calc · prerequisite — Assume integer quotient/remainder arithmetic, nonnegative integers and injectivity; at most countable means admitting an injection into N0.
• mko_definition · cited — This slice uses complete finite-set certificates or universal injectivity certificates from integer affine encodings. A failed encoding refutes that particular witness, not countability of the set. · mko-at-most-countable-set

Main derivation

(n,j)\in X

n\ge0,\quad j\in\{0,1,2\}\Rightarrow3n+j\in N_0

c(n,j)=c(m,k)

3(n-m)=k-j

j,k\in\{0,1,2\}

-2\le k-j\le2,\quad3\mid(k-j)\Rightarrow k-j=0

3(n-m)=0,\quad j=k

n=m,\quad(n,j)=(m,k)

c:X\hookrightarrow N_0

X\text{ is at most countable}

r\in N_0

r=3q+j,\quad q\in N_0,\quad j\in\{0,1,2\}

Y=\{a,b,c\},\quad e(a)=0,e(b)=1,e(c)=2

e:Y\hookrightarrow N_0,\quad |Y|=3

b(n,j)=3n

b(0,0)=b(0,1)=0,\quad c(0,0)=0\ne1=c(0,1)

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · G1, S05 — One bad code cannot imply uncountability when a valid full code already exists.
• S07 · premise_audit · G1 — Y is at most countable but not countably infinite, distinguishing the terms.
• S06 · premise_audit · G1, S04 — This extra onto certificate permits countable infinitude here; the injection alone established only at most countable.
• S05 · premise_audit · G1, S01, S04 — Surjectivity is not needed for this conclusion; finite sets are at most countable too.
• S04 · premise_audit · G1, S02, S03 — Equal images force equality of both coordinates, not just the first.
• S03 · premise_audit · G1, S02 — Both endpoints -2 and 2 are included; neither 3 nor -3 occurs.
• S02 · premise_audit · G1, S01 — No n or m is sampled; this is a universal collision analysis.
• S01 · premise_audit · G1 — Allowing negative n could leave N0, so the parameter domain matters.

Error controls and negative controls

Parameter domain

The proof fixes n≥0 and 0≤j≤2; allowing j=3 collides with the next track.

Witness versus set

Failure refutes that function, not countability of the set.

• b(n,j)=3n is injective. → reject — (0,0) and (0,1) both map to 0.
• Failure of b means X is uncountable. → reject — c(n,j)=3n+j is already universally injective.
• Finite Y is at most countable, hence countably infinite. → reject — Y has three elements and cannot biject with infinite N0.

Result and limitations

N0 times {0,1,2} injects into N0 by c(n,j)=3n+j and is at most countable.

Verification and source binding

1 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: bcd07b59b5e83e1216ba1fb8c780353d2a97cf3c8fd58ec701dee7656e7f25d5
• original source SHA-256: bcd07b59b5e83e1216ba1fb8c780353d2a97cf3c8fd58ec701dee7656e7f25d5
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

Same-source full JSON (formulas, ASTs, controls and revision log)