Main derivation
S01 · definition
Before
(X,\oplus,\odot)\ \text{finite unital ring}After
(X,\oplus)\ \text{abelian group};\ \odot\ \text{associative};\ \text{left and right distributive};\ \exists!\,1,\ 1\neq 0Conditions: none
Rule: mko_ring_def · Uses: G1, G2
Leaf basis: mko_ring_def + def_unfold
List every obligation from the MKO: addition is an abelian group (closure, associativity, identity 0, inverses, commutativity); multiplication is associative; left and right distributivity are checked separately; there is a unique two-sided multiplicative identity 1, and 1≠0. Commutativity of multiplication is not an obligation.
S02 · implies
Before
\text{the 32 entries of the two tables}After
a\oplus b=r(a+b),\ a\odot b=r(ab)\ \text{for all}\ a,b\in XConditions: none
Rule: finite_forall · Uses: G1, G2, G3, finite_forall
Leaf basis: finite_forall + division_algorithm + numeric_evaluation
Compare the 16+16 table entries with the formulas cell by cell. For example row 3 of ⊙: 3·0=0, 3·1=3, 3·2=6 with remainder 2, 3·3=9 with remainder 1, matching 0,3,2,1 in the table; row 2: 0, 2, 4 with remainder 0, 6 with remainder 2; row 3 of ⊕: 3, 4 with remainder 0, 5 with remainder 1, 6 with remainder 2. All 32 cells agree, so the two table operations are exactly addition and multiplication mod 4.
S03 · theorem_application
Before
m,\ k\in\mathbb{Z}After
r(m+k)=r(r(m)+r(k)),\ r(mk)=r(r(m)r(k))
Conditions: none
Rule: L_mod_compat · Uses: G3
Leaf basis: L_mod_compat
Lemma L_mod_compat (proved in S03.1 to S03.3) says that reducing, operating and reducing again gives the same as operating and then reducing. Every axiom below is reduced to the corresponding identity of integers.
S04 · implies
Before
a,b\in X
After
a\oplus b\in X,\ a\odot b\in X,\ r(a)=a
Conditions: none
Rule: division_algorithm · Uses: S02, G3
Leaf basis: division_algorithm
A remainder satisfies 0≤r<4, so every result lies in X: both operations are closed binary operations on X. Also, for a∈X, a=4·0+a with 0≤a<4, so r(a)=a by uniqueness; this is used below to replace c by r(c).
S05 · equivalent
Before
(a\oplus b)\oplus c=r(r(a+b)+c)
After
r(a+b+c)=a\oplus(b\oplus c),\quad a\oplus b=b\oplus a
Conditions: a,b,c\in X
Rule: L_mod_compat · Uses: S02, S03, S04
Leaf basis: L_mod_compat + add_assoc + add_comm + forall_intro
Since c=r(c), r(r(a+b)+c)=r(r(a+b)+r(c))=r((a+b)+c) by the lemma. Likewise a⊕(b⊕c)=r(a+r(b+c))=r(a+(b+c)). Integer addition is associative, so the two agree: ⊕ is associative. Also r(a+b)=r(b+a), so ⊕ is commutative. a, b, c are arbitrary elements of X, so all 64 triples are covered.
S06 · implies
Before
a\oplus 0,\ a\oplus(\ominus a)
After
a\oplus 0=a;\ \ominus0=0,\ \ominus1=3,\ \ominus2=2,\ \ominus3=1
Conditions: none
Rule: group_axioms · Uses: S02, S04, G1
Leaf basis: group_axioms + finite_forall + numeric_evaluation
a⊕0=r(a)=a and 0⊕a=a, so 0 is the additive identity (row 0 and column 0 of the addition table are 0,1,2,3). Inverses: 1⊕3=r(4)=0, 2⊕2=r(4)=0, 3⊕1=0, 0⊕0=0; each row of the addition table contains exactly one 0. So every element has an additive inverse.
S07 · implies
Before
\text{closure, associativity, identity 0, inverses, commutativity}After
(X,\oplus)\ \text{abelian group}Conditions: none
Rule: group_axioms · Uses: S04, S05, S06
Leaf basis: group_axioms + and_intro_elim
Combining S04 (closure), S05 (associativity, commutativity) and S06 (identity, inverses), (X,⊕) is an abelian group with additive identity 0.
S08 · equivalent
Before
(a\odot b)\odot c=r(r(ab)\,c)
After
r(abc)=a\odot(b\odot c)
Conditions: a,b,c\in X
Rule: L_mod_compat · Uses: S02, S03, S04
Leaf basis: L_mod_compat + mul_assoc + forall_intro
Since c=r(c), r(r(ab)c)=r(r(ab)r(c))=r((ab)c) by the lemma; a⊙(b⊙c)=r(a·r(bc))=r(r(a)r(bc))=r(a(bc)). Integer multiplication is associative, so ⊙ is associative, for every triple.
S09 · equivalent
Before
a\odot(b\oplus c)=r(a\,r(b+c))
After
r(ab+ac)=(a\odot b)\oplus(a\odot c)
Conditions: a,b,c\in X
Rule: L_mod_compat · Uses: S02, S03, S04
Leaf basis: L_mod_compat + distributive + forall_intro
Left distributivity: r(a·r(b+c))=r(r(a)r(b+c))=r(a(b+c))=r(ab+ac)=r(r(ab)+r(ac))=(a⊙b)⊕(a⊙c). The second equality and the last use the multiplicative and additive parts of the lemma; the middle uses integer distributivity.
S10 · equivalent
Before
(a\oplus b)\odot c=r(r(a+b)\,c)
After
r(ac+bc)=(a\odot c)\oplus(b\odot c)
Conditions: a,b,c\in X
Rule: L_mod_compat · Uses: S02, S03, S04
Leaf basis: L_mod_compat + distributive + forall_intro
The right law is proved separately: r(r(a+b)c)=r(r(a+b)r(c))=r((a+b)c)=r(ac+bc)=r(r(ac)+r(bc))=(a⊙c)⊕(b⊙c). It uses integer right distributivity (a+b)c=ac+bc and does not borrow commutativity of ⊙.
S11 · implies
Before
1\odot a,\ a\odot 1
After
1\odot a=a=a\odot 1\ \text{for all}\ a\in XConditions: none
Rule: mul_one · Uses: S02, S04, G2
Leaf basis: mul_one + finite_forall
1⊙a=r(a)=a and a⊙1=r(a)=a; row 1 and column 1 of the multiplication table are both 0,1,2,3. So 1 is a two-sided multiplicative identity (the MKO requires it to act on both sides).
S12 · theorem_application
Before
e\ \text{a two-sided identity for}\ \odotAfter
e=1
Conditions: none
Rule: L_unique_identity · Uses: S11
Leaf basis: L_unique_identity
By Lemma L_unique_identity (proved in S12.1), any two-sided identity e equals 1, so the multiplicative identity is unique. The table agrees: only row 1 equals 0,1,2,3.
S13 · implies
Before
0\ \text{additive identity},\ 1\ \text{multiplicative identity}After
1\neq 0
Conditions: none
Rule: numeric_evaluation · Uses: S06, S11
Leaf basis: numeric_evaluation + mul_one
1 and 0 are different elements of X, and they act differently: 1⊙1=1 while 0⊙1=0. The MKO requires 1≠0, which rules out the one-element zero ring.
S14 · implies
Before
\text{all obligations of S01}After
(X,\oplus,\odot)\ \text{is a finite unital ring};\ 2\odot2=0Conditions: none
Rule: mko_ring_def · Uses: S01, S07, S08, S09, S10, S11, S12, S13
Leaf basis: mko_ring_def + and_intro_elim
Every obligation of S01 is proved: S07 (additive abelian group), S08 (associative multiplication), S09 and S10 (both distributive laws), S11 to S13 (unique two-sided identity with 1≠0). So it is a finite unital ring. Also 2⊙2=r(4)=0 with 2≠0, so 2 is a zero divisor: the ring is not an integral domain or a field, but the MKO does not require those (they are separate nodes).