#!/usr/bin/env python3
# SMT proof (z3) of the DISTINCT-KEY THRESHOLD property for AereThresholdAccount +
# AereThresholdPQCRegistry, formalising the duplicate-key finding fixed 2026-07-15.
#
# THE BUG: authorization counts distinct signers by member INDEX (the `seen` bitmask
# in _countMem / the strict DuplicateMember check). Member indices are 0..n-1, so they
# are ALWAYS distinct -- the index dedup can never fail on a real committee. Distinctness
# of the actual signing KEYS was never checked at committee registration, so the same
# public key could occupy two indices and one keyholder could fill multiple "distinct
# member" slots and reach the threshold t alone, collapsing the t-of-n guarantee.
#
# THE FIX: reject duplicate committee keys at the one place a committee is set
# (initialize / registerCommittee). Modelled here as Distinct(key).
#
# MODEL: key[i] = the id of the KEYHOLDER controlling slot i. A single keyholder h
# occupies slot i iff key[i] == h; the number of slots it can validly sign for is
# count(i: key[i] == h). "One keyholder reaches threshold" := exists h with count >= t.
#
# METHOD: assert the property's NEGATION; UNSAT => the property holds.

from z3 import Int, Solver, Sum, If, Distinct, sat, unsat

results = []

def check(name, s, expect_unsat=True, kind="PROOF"):
    r = s.check()
    ok = (r == unsat) if expect_unsat else (r == sat)
    tag = "PROVED" if (ok and expect_unsat) else ("CEX-FOUND" if (ok and not expect_unsat) else "FAILED")
    results.append((name, tag, ok, kind))
    print(f"[{tag}] ({kind}) {name}: z3={r}")
    if r == sat and not expect_unsat:
        m = s.model()
        print(f"             witness: {m}")
    return ok

def keys(N):
    return [Int(f"key_{i}") for i in range(N)]

def count_for(key, h):
    return Sum([If(key[i] == h, 1, 0) for i in range(len(key))])

print("### AereThresholdAccount / PQC registry -- DISTINCT-KEY THRESHOLD (dup-key finding 2026-07-15)\n")
print("    Distinct signers are counted by member INDEX (0..n-1, always distinct), so the")
print("    index dedup never fires on a real committee. The guarantee that matters is that")
print("    the KEYS are distinct. Proved below; the pre-fix (no key-distinctness) is CEX'd.\n")

print("=" * 74)
print("P1  distinct committee keys => every keyholder fills at most 1 slot, so reaching")
print("    threshold t>=2 requires >= t DISTINCT keyholders (the t-of-n guarantee holds)")
print("=" * 74)
for (N, T) in [(5, 2), (5, 3), (9, 5), (7, 4), (21, 11)]:
    s = Solver()
    key = keys(N)
    h = Int("h")
    s.add(Distinct(key))            # THE FIX: committee keys are pairwise distinct
    s.add(count_for(key, h) >= T)   # NEGATION: some single keyholder h fills >= t slots
    check(f"N={N} t={T}: distinct keys AND one keyholder fills >= t is impossible", s, expect_unsat=True)

print("\n" + "=" * 74)
print("P2  NEG-CONTROL: WITHOUT the distinct-key check, one keyholder CAN reach the")
print("    threshold alone (the exact pre-fix bug: a committee like [A,A,A,B,C], t=3)")
print("=" * 74)
for (N, T) in [(5, 3), (9, 5)]:
    s = Solver()
    key = keys(N)
    h = Int("h")
    # No Distinct(key): a keyholder may occupy multiple indices (duplicate keys allowed).
    s.add(count_for(key, h) >= T)
    check(f"N={N} t={T}: no key-distinctness => one keyholder filling >= t is POSSIBLE",
          s, expect_unsat=False, kind="NEG-CONTROL")

print("\n" + "=" * 74)
print("P3  index-distinctness ALONE does not imply key-distinctness (why the seen-bitmask")
print("    dedup was insufficient): indices are trivially distinct, keys can still repeat")
print("=" * 74)
for (N, T) in [(5, 3)]:
    s = Solver()
    key = keys(N)
    h = Int("h")
    idx = [Int(f"idx_{i}") for i in range(N)]
    s.add(Distinct(idx))                         # the index dedup the contract DOES have
    s.add([idx[i] == i for i in range(N)])       # indices are 0..n-1 (always satisfiable)
    s.add(count_for(key, h) >= T)                # yet one keyholder still fills >= t slots
    check(f"N={N} t={T}: indices distinct yet one keyholder fills >= t (key dedup needed)",
          s, expect_unsat=False, kind="NEG-CONTROL")

print("\n=== SUMMARY (distinct-key threshold, AereThresholdAccount / PQC registry) ===")
allok = True
for name, tag, ok, kind in results:
    print(f"  {tag:9} [{kind}]  {name}")
    allok = allok and ok
print()
if allok:
    print("  ESTABLISHED: the 2026-07-15 fix (reject duplicate committee keys at registration)")
    print("  is exactly what makes t-of-n sound. P1 PROVES that with distinct keys no single")
    print("  keyholder can reach threshold t>=2, so t signatures require t distinct parties.")
    print("  P2/P3 are load-bearing neg-controls: without key-distinctness (index dedup alone),")
    print("  one keyholder reaches the threshold alone -- the collapsed-guarantee bug that was")
    print("  found by manual review and fixed + redeployed (factory V2, PQC registry V2).")
    print("  BOUNDARY: a combinatorial model of the counting logic, not the Solidity bytecode;")
    print("  complements the on-chain proof that the live V2 reverts DuplicatePubKey.")
else:
    print("  NOT fully established (see FAILED above).")

import sys
sys.exit(0 if allok else 1)