Main derivation
S01 · definition
Before
d=\gcd(a,b)
After
d\mid a\ \land\ d\mid b\ \land\ \forall c\in\mathbb{N}\,(c\mid a\land c\mid b\Rightarrow c\mid d)Conditions: none
Rule: mko_gcd_def · Uses: G1, mko_gcd_def
Leaf basis: mko_gcd_def + def_unfold
By the MKO, the gcd is characterized in two parts: d is a common divisor, and every common divisor divides d. 'Greatest' is expressed by divisibility, not by numerical size. Coprime means gcd=1.
S02 · implies
Before
d,\ d'\ \text{both satisfy S01 for}\ (a,b)After
d=d'
Conditions: none
Rule: L_dvd_antisymm · Uses: S01, L_dvd_antisymm
Leaf basis: mko_dvd_def + exists_elim + order_basics
d′ is a common divisor, and the characterization of d says every common divisor divides d, so d′∣d; symmetrically d∣d′. By Lemma L_dvd_antisymm (proved in S02.1 and S02.2), d=d′. So whenever it exists, the gcd is unique and may be called 'the' gcd.
S03 · implies
Before
(x,0)
After
x\ \text{satisfies S01 for}\ (x,0),\ \text{so}\ \gcd(x,0)=xConditions: none
Rule: dvd_basic · Uses: S01, dvd_basic
Leaf basis: dvd_basic + and_intro_elim
x∣x and x∣0 (dvd_basic), so x is a common divisor; if c∣x and c∣0, then certainly c∣x. So x satisfies the characterization, and by S02 it is gcd(x,0). This includes x=0: 0 satisfies the characterization for (0,0), so the MKO's convention gcd(0,0)=0 agrees with the characterization.
S04 · equivalent
Before
x=yq+r,\ y>0
After
c\mid x\land c\mid y\iff c\mid y\land c\mid r
Conditions: y>0
Rule: L_cd_step · Uses: mko_dvd_def, L_cd_step
Leaf basis: distributive + order_basics + exists_intro
By Lemma L_cd_step (proved in S04.1 and S04.2), one step of the algorithm does not change the set of common divisors: CD(x,y)=CD(y,r).
S05 · equivalent
Before
x=yq+r,\ y>0
After
d\ \text{satisfies S01 for}\ (x,y)\iff d\ \text{satisfies S01 for}\ (y,r)Conditions: y>0
Rule: set_extensionality · Uses: S01, S04
Leaf basis: set_extensionality + eq_replace
The characterization S01 depends on the inputs only through the set of common divisors: 'd∈CD and every c in CD divides d'. S04 says CD(x,y)=CD(y,r), so the two characterizations are identical. Hence gcd(x,y)=gcd(y, x mod y).
S06 · implies
Before
(84,36)
After
84=36\cdot 2+12,\ 0\le 12<36;\ (84,36)\mapsto(36,12)
Conditions: 36>0
Rule: division_algorithm · Uses: G2, G3, division_algorithm
Leaf basis: division_algorithm + numeric_evaluation
First step: 36≠0, and 84 divided by 36 gives quotient 2 and remainder 12 (N1, N2), so (84,36) becomes (36,12).
S07 · implies
Before
(36,12)
After
36=12\cdot 3+0;\ (36,12)\mapsto(12,0)
Conditions: 12>0
Rule: division_algorithm · Uses: S06, division_algorithm
Leaf basis: division_algorithm + numeric_evaluation
Second step: 12≠0 and 36=12·3+0 (N3), so (36,12) becomes (12,0). The second entry is 0, so the algorithm stops and outputs 12.
S08 · implies
Before
\gcd(12,0)=12
After
\gcd(84,36)=\gcd(36,12)=\gcd(12,0)=12
Conditions: none
Rule: logic_chain · Uses: S03, S05, S06, S07
Leaf basis: logic_chain + eq_refl_symm_trans
By S03, 12 is gcd(12,0). S05 applied to the step of S07 (36=12·3+0) shows 12 also satisfies the characterization for (36,12); applied again to the step of S06 (84=36·2+12) it shows 12 satisfies the characterization for (84,36). So gcd(84,36)=12 (N4 is an independent recomputation).
S09 · implies
Before
(x,y)\mapsto(y,\,x\bmod y),\ y>0
After
\text{the algorithm stops after finitely many steps}Conditions: none
Rule: well_ordering · Uses: G2, division_algorithm, well_ordering
Leaf basis: well_ordering + division_algorithm + contradiction
Each step replaces the second entry y by the remainder r with 0≤r<y. If the algorithm never stopped, the second entries would form an infinite strictly decreasing sequence of naturals; its set of values is nonempty but has no least element (every value is followed by a smaller one), contradicting well-ordering. So the algorithm reaches second entry 0 after finitely many steps, at most the initial y of them. Here the second entries 36, 12, 0 decrease strictly.
S10 · implies
Before
d=12
After
12\mid 84,\ 12\mid 36,\ (c\mid 84\land c\mid 36\Rightarrow c\mid 12)
Conditions: none
Rule: mko_gcd_def · Uses: S04, S06, S08
Leaf basis: mko_gcd_def + exists_intro + numeric_evaluation
Verify both parts of the characterization directly: 84=12·7 and 36=12·3 (N5, N6), so 12 is a common divisor; for any common divisor c, the ⇒ direction of S04 (x=84, y=36, q=2, r=12) gives c∣12. For example the common divisors 1, 2, 3, 4, 6, 12 of 84 and 36 all divide 12.
S11 · implies
Before
6\mid 84,\ 6\mid 36
After
12\nmid 6,\ \text{so}\ 6\neq\gcd(84,36)Conditions: none
Rule: mko_gcd_def · Uses: S01, S10
Leaf basis: mko_gcd_def + order_basics + numeric_evaluation
84=6·14 and 36=6·6, so 6 is a common divisor. But 12 is also a common divisor and 12∤6: in 6=12·k, k=0 gives 0 and k≥1 gives at least 12, neither equal to 6. The second part of the characterization fails, so 6 is not the gcd. This is the misconception the MKO warns about: calling any common divisor the greatest one.
S12 · implies
Before
(8,15)
After
(8,15)\mapsto(15,8)\mapsto(8,7)\mapsto(7,1)\mapsto(1,0);\ \gcd(8,15)=1
Conditions: none
Rule: division_algorithm · Uses: G2, G3, S03, S05
Leaf basis: division_algorithm + logic_chain + numeric_evaluation
Step by step: 8=15·0+8 →(15,8); 15=8·1+7 →(8,7); 8=7·1+1 →(7,1); 7=1·7+0 →(1,0) (N10 to N13). Every divisor used is positive. By S03, gcd(1,0)=1, and applying S05 backwards step by step, gcd(8,15)=1 (N14 is an independent recomputation). The first step only swaps the order, since 8<15.
S13 · implies
Before
\gcd(8,15)=1,\ 8=2\cdot 4,\ 15=3\cdot 5
After
8,15\ \text{coprime, neither prime}Conditions: none
Rule: mko_gcd_def · Uses: S12, mko_gcd_def
Leaf basis: mko_gcd_def + exists_intro + numeric_evaluation
By the definition, gcd=1 means coprime. 8 has the divisor 2 (2≠1, 2≠8) and 15 has the divisor 3 (3≠1, 3≠15), so neither is prime. Coprimality is a relation between two numbers and does not require either to be prime (MKO paragraph 2).
S14 · implies
Before
(0,1),\ (0,6),\ (0,0)
After
\gcd(0,1)=1,\ \gcd(0,6)=6,\ \gcd(0,0)=0
Conditions: none
Rule: logic_chain · Uses: G2, G3, S03, S05
Leaf basis: logic_chain + dvd_basic + numeric_evaluation
(0,1): 0=1·0+0 →(1,0), gcd=1, so 0 and 1 are coprime. (0,6): 0=6·0+0 →(6,0), gcd=6≠1, not coprime (2 divides both 0 and 6). (0,0): by S03, 0 satisfies the characterization, agreeing with the convention; note that every natural divides 0, so (0,0) has no numerically largest common divisor, and the divisibility characterization is what makes 0 the natural answer.