Use exact integer n=3>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).
G001 — Use exact integer n=3>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).
Definition and scope
Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.
Wilson ORIGINAL factorial ORIGINAL running factorial product1 := mul(1, 1)
After
Wilson ORIGINAL factorial ORIGINAL running factorial product1 = 1
Conditions: none
Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate.
source roles = [Wilson ORIGINAL factorial, 3, 1, 1, 1]
After
derived = [0, 1, Yes, Yes]
Conditions: none
Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry.
Wilson ORIGINAL factorial ORIGINAL running factorial product2 := mul(1, 2)
After
Wilson ORIGINAL factorial ORIGINAL running factorial product2 = 2
Conditions: none
Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate.
source roles = [Wilson ORIGINAL factorial, 3, 2, 1, 2]
After
derived = [0, 2, Yes, Yes]
Conditions: none
Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry.
Qualify primality and congruence separately.Both may be False at a composite while equivalence is True.The converse has a separate argument:a proper divisor d>=2 of composite n divides both n and(n-1)!;a congruence to-1 would force d to divide1,a contradiction.This reasoning is not a single instance or finite-cutoff test.
S006 · premise_and_result_audit — Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct.
S005 · premise_and_result_audit — Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct.
S004 · premise_and_result_audit — Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product2 with exact rational arithmetic, not rounded decimals.
S003 · premise_and_result_audit — Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct.
S002 · premise_and_result_audit — Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product1 with exact rational arithmetic, not rounded decimals.
S001 · premise_and_result_audit — Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct.
The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.
This question review and source binding
Review date: 2026-10-26
Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.