Main derivation
S01 · definition
Before
d(x,y)=\|x-y\|
After
d\ge0,\ d(x,y)=0\Leftrightarrow x=y,\ d(x,y)=d(y,x),\ d(x,z)\le d(x,y)+d(y,z)
Conditions: none
Rule: mko_nim_def · Uses: G1, G2, metric_def
Leaf basis: mko_nim_def + metric_def + def_unfold
By the MKO and the definition of a metric, four properties must be proved, each for all x, y, z∈V: nonnegativity, separation, symmetry and the triangle inequality. The proof uses only the three norm axioms of G1, no particular points.
S02 · implies
Before
x,y\in V
After
x-y=x+(-1)y\in V,\ d:V\times V\to\mathbb{R}Conditions: none
Rule: vector_space_axioms · Uses: G1, G2, vector_space_axioms
Leaf basis: vector_space_axioms + function_def
A vector space is closed under addition and scalar multiplication, so x−y=x+(−1)y lies in V and ‖x−y‖ is a real number. d is a function V×V→ℝ. This needs x and y in the same space (assumption a1).
S03 · implies
Before
d(x,y)=\|x-y\|
After
d(x,y)\ge 0
Conditions: none
Rule: mko_norm_def · Uses: S02, G1
Leaf basis: mko_norm_def
N1 applied to the vector v=x−y gives ‖x−y‖≥0, so d(x,y)≥0.
S04 · equivalent
Before
d(x,y)=0
After
x=y
Conditions: none
Rule: mko_norm_def · Uses: S02, G1
Leaf basis: mko_norm_def + vector_space_axioms
d(x,y)=0 ⇔ ‖x−y‖=0 ⇔ x−y=0 (second half of N1) ⇔ x=y (S04.1). This is separation: the distance is zero exactly when the points coincide.
S05 · equivalent
Before
d(x,y)=\|x-y\|
After
d(x,y)=\|y-x\|=d(y,x)
Conditions: none
Rule: mko_norm_def · Uses: S02, G1
Leaf basis: mko_norm_def + vector_space_axioms + abs_rules
Symmetry comes from two sub-steps: S05.1 writes x−y as (−1)(y−x), and S05.2 uses N2 with t=−1 to get ‖(−1)(y−x)‖=‖y−x‖.
S06 · implies
Before
d(x,z)=\|x-z\|
After
d(x,z)\le d(x,y)+d(y,z)
Conditions: none
Rule: mko_norm_def · Uses: S02, G1
Leaf basis: mko_norm_def + vector_space_axioms
The triangle inequality comes from two sub-steps: S06.1 splits x−z as (x−y)+(y−z), and S06.2 applies N3 to this sum.
S07 · implies
Before
\text{nonnegativity, separation, symmetry, triangle}After
d(x,y)=\|x-y\|\ \text{is a metric on}\ VConditions: none
Rule: metric_def · Uses: S03, S04, S05, S06, metric_def
Leaf basis: metric_def + forall_intro + and_intro_elim
All four properties are proved for arbitrary x, y, z with no extra assumption on them, so by generalization and the definition of a metric, d is a metric on V. This is a universal proof using no finite sample.
S08 · theorem_application
Before
\|v\|_1=\sum_{i=1}^{n}|v_i|After
\|\cdot\|_1\ \text{satisfies N1, N2, N3}Conditions: none
Rule: L_l1_norm · Uses: G3
Leaf basis: L_l1_norm
Lemma L_l1_norm (proved in S08.1 to S08.3): the l1 formula applies the properties of absolute value to each coordinate and adds over the finite sum, giving the three norm axioms. This is an algebraic certificate for all vectors (the qualification of the named norm in the MKO).
S09 · theorem_application
Before
\|v\|_\infty=\max_i|v_i|
After
\|\cdot\|_\infty\ \text{satisfies N1, N2, N3}Conditions: none
Rule: L_linf_norm · Uses: G3
Leaf basis: L_linf_norm
Lemma L_linf_norm (proved in S09.1 to S09.3): the l∞ formula satisfies the norm axioms by the properties of a finite maximum.
S10 · theorem_application
Before
\|\cdot\|_1,\ \|\cdot\|_\infty\ \text{norms on}\ \mathbb{R}^nAfter
d_1(x,y)=\|x-y\|_1,\ d_\infty(x,y)=\|x-y\|_\infty\ \text{are metrics}Conditions: none
Rule: mko_nim_def · Uses: S07, S08, S09
Leaf basis: mko_nim_def + L_l1_norm + L_linf_norm
S08 and S09 confirm that both are norms on ℝⁿ, so (ℝⁿ,‖·‖₁) and (ℝⁿ,‖·‖∞) are normed spaces, and the general theorem S07 gives the two metrics d1 and d∞.
S11 · implies
Before
x-y=(1-4,\,-2-0,\,3-(-1))=(-3,-2,4)
After
d_1(x,y)=3+2+4=9,\ d_\infty(x,y)=\max(3,2,4)=4
Conditions: none
Rule: numeric_evaluation · Uses: S10, G3
Leaf basis: numeric_evaluation + abs_rules
Subtract coordinatewise: x−y=(−3,−2,4), with absolute values 3, 2, 4. l1 takes the sum, 9, and l∞ the maximum, 4.
S12 · implies
Before
y-x=(3,2,-4)
After
d_1(y,x)=9=d_1(x,y)
Conditions: none
Rule: numeric_evaluation · Uses: S05, S11, G3
Leaf basis: numeric_evaluation + abs_rules
y−x=(3,2,−4) also has absolute values 3, 2, 4, so d1(y,x)=9, consistent with the general symmetry of S05. This is one example, not a proof of symmetry.
S13 · implies
Before
x-z=(-1,-7,2),\ y-z=(2,-5,-2)
After
d_1(x,z)=10,\ d_\infty(x,z)=7,\ d_1(y,z)=9,\ d_\infty(y,z)=5
Conditions: none
Rule: numeric_evaluation · Uses: S10, G3
Leaf basis: numeric_evaluation + abs_rules
x−z=(1−2,−2−5,3−1)=(−1,−7,2): l1 gives 1+7+2=10 and l∞ gives 7. y−z=(4−2,0−5,−1−1)=(2,−5,−2): l1 gives 2+5+2=9 and l∞ gives 5.
S14 · implies
Before
d_1(x,z)=10,\ d_\infty(x,z)=7
After
10\le 9+9,\ 7\le 4+5
Conditions: none
Rule: order_basics · Uses: S06, S11, S13
Leaf basis: order_basics + numeric_evaluation
At these three points the triangle inequality reads 10≤18 for d1 and 7≤9 for d∞, both consistent with S06. This checks an example and cannot replace the universal proof of S06.
S15 · implies
Before
d_1(x,y)=9,\ d_\infty(x,y)=4
After
d_1(x,y)\neq d_\infty(x,y)
Conditions: none
Rule: numeric_evaluation · Uses: S11
Leaf basis: numeric_evaluation
The same pair of points has different distances under the two named norms (9≠4), so each computation must identify its norm and keep both points in the same dimension (assumption a1). A mixed-norm bound needs an additional comparison estimate, not just the triangle inequality for one metric. Here ‖v‖∞≤‖v‖₁ gives d∞(x,z)≤d1(x,y)+d∞(y,z).
Substeps and lemmas
S04.1 · equivalent
Before
x-y=0
After
x=y
Conditions: none
Rule: vector_space_axioms · Uses: vector_space_axioms
Leaf basis: vector_space_axioms + eq_congruence
Add y to both sides: (x−y)+y=x+((−y)+y)=x+0=x, and on the right 0+y=y, so x=y. Backward: if x=y then x−y=y−y=0.
S05.1 · equivalent
Before
x-y
After
(-1)(y-x)
Conditions: none
Rule: vector_space_axioms · Uses: vector_space_axioms
Leaf basis: vector_space_axioms + neg_mul
(−1)(y−x)=(−1)y+(−1)(−1)x=−y+x=x−y, using distributivity of a scalar over vector addition, (ab)u=a(bu) and commutativity of addition.
S05.2 · equivalent
Before
\|(-1)(y-x)\|
After
|-1|\,\|y-x\|=\|y-x\|
Conditions: t=-1
Rule: mko_norm_def · Uses: S05.1, G1
Leaf basis: mko_norm_def + abs_rules + mul_one
N2 with t=−1: ‖(−1)(y−x)‖=|−1|·‖y−x‖=‖y−x‖. This step really needs N2; a function satisfying only N1 and N3 need not give a symmetric distance (see the error controls).
S06.1 · equivalent
Before
x-z
After
(x-y)+(y-z)
Conditions: none
Rule: vector_space_axioms · Uses: vector_space_axioms
Leaf basis: vector_space_axioms
(x−y)+(y−z)=x+((−y)+y)+(−z)=x+0−z=x−z, using associativity, inverses and the zero vector.
S06.2 · implies
Before
\|(x-y)+(y-z)\|
After
\le\|x-y\|+\|y-z\|
Conditions: none
Rule: mko_norm_def · Uses: S06.1, G1
Leaf basis: mko_norm_def
N3 with u=x−y and v=y−z: ‖u+v‖≤‖u‖+‖v‖. Combined with S06.1, d(x,z)≤d(x,y)+d(y,z).
S08.1 · implies
Before
\sum_{i}|v_i|After
\sum_i|v_i|\ge0,\ \sum_i|v_i|=0\Leftrightarrow v=0
Conditions: none
Rule: abs_rules · Uses: G3, abs_rules, order_basics
Leaf basis: abs_rules + order_basics + finite_sum_rules
Each |v_i|≥0, and a sum of nonnegative numbers is nonnegative. If the sum is 0, each term lies between 0 and the whole sum, so every |v_i|=0, that is every v_i=0 and v=0; conversely v=0 gives the sum 0. This is N1.
S08.2 · equivalent
Before
\sum_i|t\,v_i|
After
|t|\sum_i|v_i|
Conditions: none
Rule: abs_rules · Uses: G3, abs_rules, finite_sum_rules
Leaf basis: abs_rules + finite_sum_rules
|t v_i|=|t||v_i|, then the constant |t| comes out of the finite sum. This is N2.
S08.3 · implies
Before
\sum_i|u_i+v_i|
After
\le\sum_i(|u_i|+|v_i|)=\sum_i|u_i|+\sum_i|v_i|
Conditions: none
Rule: abs_rules · Uses: G3, abs_rules, order_basics
Leaf basis: abs_rules + order_basics + finite_sum_rules
Each coordinate satisfies |u_i+v_i|≤|u_i|+|v_i| (the real triangle inequality). Adding the n inequalities (u<v gives u+w<v+w, used repeatedly, allowing equality) gives the inequality of sums, and the sum then splits. This is N3. S08.1 to S08.3 together prove Lemma L_l1_norm.
S09.1 · implies
Before
\max_i|v_i|
After
\max_i|v_i|\ge0,\ \max_i|v_i|=0\Leftrightarrow v=0
Conditions: none
Rule: max_rules · Uses: G3, abs_rules, max_rules
Leaf basis: max_rules + abs_rules + order_basics
The maximum equals some |v_k|≥0. If the maximum is 0, every |v_i|≤0, and also |v_i|≥0, so every v_i=0; conversely v=0 gives maximum 0. This is N1.
S09.2 · equivalent
Before
\max_i|t\,v_i|
After
|t|\max_i|v_i|
Conditions: |t|\ge 0
Rule: max_rules · Uses: G3, abs_rules, max_rules
Leaf basis: max_rules + abs_rules + order_basics
|t v_i|=|t||v_i|. Let the maximum be attained at k: for every i, |v_i|≤|v_k|, and multiplying by the nonnegative |t| keeps the inequality, so |t||v_i|≤|t||v_k|; the maximum is |t||v_k|=|t|max|v_i|. This is N2.
S09.3 · implies
Before
|u_i+v_i|\le|u_i|+|v_i|\le\|u\|_\infty+\|v\|_\infty
After
\|u+v\|_\infty\le\|u\|_\infty+\|v\|_\infty
Conditions: none
Rule: max_rules · Uses: G3, abs_rules, max_rules
Leaf basis: abs_rules + max_rules + order_basics
For every i: |u_i+v_i|≤|u_i|+|v_i| (the real triangle inequality), with |u_i|≤‖u‖∞ and |v_i|≤‖v‖∞. So ‖u‖∞+‖v‖∞ is an upper bound of every |u_i+v_i|, and the maximum does not exceed it. This is N3. S09.1 to S09.3 together prove Lemma L_linf_norm.
Error controls and negative controls
A finite sample does not establish the metric or its origin
The common mistake named by the MKO is to claim, after checking finitely many distances, that a metric is induced by a norm. The 10≤18 and 7≤9 of S14 are only examples. Counterexample: the discrete metric on ℝ³ (distance 1 between distinct points) satisfies the metric axioms, but a norm-induced metric always has d(2v,0)=‖2v‖=2d(v,0), while the discrete metric gives 1≠2 for v≠0, so no norm induces it.
N1 and N3 alone do not ensure symmetry
N1 and N3 alone are not enough. On ℝ, f(v)=max(v,−2v) is nonnegative, zero only at 0, and subadditive (a maximum of linear functions), but f(1)=1 and f(−1)=2, so f(x−y) is not symmetric: from 1 to 0 it is 1, from 0 to 1 it is 2. S05.2 obtains symmetry exactly through N2 with t=−1.
One named norm, one dimension
d1(x,y)=9 while d∞(x,y)=4 (S15). Mixing norms requires a comparison estimate in addition to the triangle inequality for one metric. In this example d∞(x,y)≤d1(x,y), so d∞(x,z)≤d∞(x,y)+d∞(y,z)≤d1(x,y)+d∞(y,z) does hold. When the two points have different dimensions, x−y is undefined in this model (S02).
- d(x,y)=‖x‖₁−‖y‖₁ is a metric on ℝ³ →
reject — x=0 and y=(1,0,0) give 0−1=−1<0, violating nonnegativity. - The discrete metric on ℝ³ is induced by some norm →
reject — An induced metric satisfies d(2v,0)=2d(v,0), but the discrete metric gives 1≠2 for v≠0. - The triangle inequality was checked at x, y, z (10≤18, 7≤9), so d1 and d∞ are metrics →
reject — A finite sample supports only those points; the metric property comes from the universal proof in S06 to S10. - The l1 distance between (1,−2,3)∈ℝ³ and (1,2)∈ℝ² is 4 →
reject — The points are not in the same space, so x−y is undefined (assumption a1).