Main derivation
S01 · definition
Before
\|\cdot\|_1,\ \|\cdot\|_\infty\ \text{are norms on}\ \mathbb{R}^nAfter
(N1)\land(N2)\land(N3)\ \text{for all } u,v\in\mathbb{R}^n,\ a\in\mathbb{R},\ \text{for each formula}Conditions: none
Rule: mko_norm · Uses: G2, G3, mko_norm
Leaf basis: mko_norm + def_unfold
By the MKO, a formula is a norm exactly when the three axioms hold for all vectors and all scalars. What is needed is therefore a universal certificate: every step argues for arbitrary u, v, a, rather than substituting a few vectors.
S02 · implies
Before
|v_i|\ge 0\ \text{for every } iAfter
\|v\|_1=\sum_{i=1}^{n}|v_i|\ge 0Conditions: none
Rule: sum_order · Uses: G2, abs_rules, sum_order
Leaf basis: abs_rules + sum_order
Absolute values are nonnegative: every |v_i|≥0. A finite sum of nonnegative numbers is nonnegative, so ‖v‖₁≥0 (the first half of N1).
S03 · equivalent
Before
\|v\|_1=0
After
v=0
Conditions: none
Rule: sum_order · Uses: S02, G1, sum_order, L_abs_zero
Leaf basis: sum_order + L_abs_zero + coord_ops
Forward: if ‖v‖₁=0, then for every j, 0≤|v_j|≤Σ|v_i|=0 (a sum of nonnegative terms is at least each term), so |v_j|=0; by Lemma L_abs_zero (S03.1), v_j=0; every coordinate is 0, that is v=0. Backward: if v=0 then every |v_i|=|0|=0 and the sum is 0. This is the second half of N1.
S04 · equivalent
Before
\|av\|_1=\sum_{i=1}^{n}|a v_i|After
\|av\|_1=\sum_{i=1}^{n}|a||v_i|=|a|\,\|v\|_1Conditions: none
Rule: finite_sum_rules · Uses: G1, G2, abs_rules, finite_sum_rules
Leaf basis: coord_ops + abs_rules + finite_sum_rules
(av)_i=a v_i and |a v_i|=|a||v_i|; the constant |a| factors out of the finite sum: Σ|a||v_i|=|a|Σ|v_i|. This holds for arbitrary a and v, which is N2.
S05 · implies
Before
\|u+v\|_1=\sum_{i=1}^{n}|u_i+v_i|After
\|u+v\|_1\le\sum_{i=1}^{n}(|u_i|+|v_i|)=\|u\|_1+\|v\|_1Conditions: none
Rule: sum_order · Uses: G1, G2, abs_rules, sum_order, finite_sum_rules
Leaf basis: abs_rules + sum_order + finite_sum_rules + coord_ops
In each coordinate use the real triangle inequality |u_i+v_i|≤|u_i|+|v_i|; adding the termwise inequalities gives the inequality for the sums; then split Σ(|u_i|+|v_i|) into Σ|u_i|+Σ|v_i|. This holds for arbitrary u and v, which is N3.
S06 · implies
Before
(N1),(N2),(N3)\ \text{for}\ \|\cdot\|_1\ \text{with arbitrary } u,v,aAfter
\|\cdot\|_1\ \text{is a norm on}\ \mathbb{R}^nConditions: none
Rule: forall_intro · Uses: S02, S03, S04, S05, forall_intro
Leaf basis: forall_intro + and_intro_elim
S02 to S05 prove the three axioms for arbitrary u, v, a, without using particular values; by ∀-introduction and conjunction, the ℓ¹ formula is a norm on ℝⁿ.
S07 · implies
Before
\|v\|_\infty=\max_i|v_i|
After
\|v\|_\infty=|v_k|\ge 0\ \text{for some } kConditions: n\ge 1
Rule: max_finite · Uses: G2, max_finite, abs_rules
Leaf basis: max_finite + abs_rules
Since n≥1, the maximum of finitely many numbers exists and equals some |v_k|, and absolute values are nonnegative, so ‖v‖_∞≥0 (the first half of N1).
S08 · equivalent
Before
\|v\|_\infty=0
After
v=0
Conditions: none
Rule: max_finite · Uses: S07, G1, max_finite, L_abs_zero
Leaf basis: max_finite + L_abs_zero + coord_ops
Forward: if max|v_i|=0, then 0≤|v_i|≤0 for every i, so |v_i|=0 and by L_abs_zero v_i=0, that is v=0. Backward: if v=0, every |v_i|=0 and the maximum is 0.
S09 · equivalent
Before
\|av\|_\infty=\max_i|a||v_i|
After
\|av\|_\infty=|a|\max_i|v_i|=|a|\,\|v\|_\infty
Conditions: |a|\ge 0
Rule: max_finite · Uses: G1, G2, abs_rules, max_finite, mono_nonneg
Leaf basis: abs_rules + max_finite + mono_nonneg + order_basics
|(av)_i|=|a||v_i|. Let M=max|v_i|=|v_k|. Upper bound (S09.1): every |a||v_i|≤|a|M, so the maximum is ≤|a|M; lower bound (S09.2): |a||v_k|=|a|M is itself one of the values, so the maximum is ≥|a|M. Together they are equal, which is N2. What is factored out is |a|≥0, not a.
S10 · implies
Before
|u_i+v_i|\ \text{for each } iAfter
\|u+v\|_\infty\le\|u\|_\infty+\|v\|_\infty
Conditions: none
Rule: max_finite · Uses: G1, G2, abs_rules, max_finite, ineq_add
Leaf basis: abs_rules + max_finite + ineq_add + coord_ops
For each i: |u_i+v_i|≤|u_i|+|v_i| (triangle inequality), while |u_i|≤‖u‖_∞ and |v_i|≤‖v‖_∞; adding gives |u_i+v_i|≤‖u‖_∞+‖v‖_∞. Every value is at most the same bound K=‖u‖_∞+‖v‖_∞, so the maximum is at most K. This is N3.
S11 · implies
Before
(N1),(N2),(N3)\ \text{for}\ \|\cdot\|_\infty\ \text{with arbitrary } u,v,aAfter
\|\cdot\|_\infty\ \text{is a norm on}\ \mathbb{R}^nConditions: none
Rule: forall_intro · Uses: S07, S08, S09, S10, forall_intro
Leaf basis: forall_intro + and_intro_elim
S07 to S10 prove the three axioms for arbitrary u, v, a; by ∀-introduction and conjunction, the ℓ^∞ formula is a norm on ℝⁿ.
S12 · implies
Before
u=(3,-4,1),\ v=(-1,2,5)
After
\|u\|_1=8,\ \|u\|_\infty=4,\ \|v\|_1=8,\ \|v\|_\infty=5
Conditions: none
Rule: numeric_evaluation · Uses: G4, G2
Leaf basis: numeric_evaluation + abs_rules
Apply the formulas: ‖u‖₁=3+4+1=8 and ‖u‖_∞=max(3,4,1)=4; ‖v‖₁=1+2+5=8 and ‖v‖_∞=max(1,2,5)=5. The two vectors have the same ℓ¹ norm but different ℓ^∞ norms: different norms measure size differently.
S13 · implies
Before
u+v=(2,-2,6)
After
\|u+v\|_1=10\le 16,\ \|u+v\|_\infty=6\le 9
Conditions: none
Rule: numeric_evaluation · Uses: S12, S05, S10
Leaf basis: numeric_evaluation + coord_ops
u+v=(2,−2,6). ‖u+v‖₁=2+2+6=10≤8+8=16; ‖u+v‖_∞=6≤4+5=9. These values agree with the general results of S05 and S10, but they are instances only, not proofs.
S14 · implies
Before
-2u=(-6,8,-2)
After
\|-2u\|_1=16=|-2|\cdot 8,\ \|-2u\|_\infty=8=|-2|\cdot 4
Conditions: none
Rule: numeric_evaluation · Uses: S12, S04, S09
Leaf basis: numeric_evaluation + coord_ops
−2u=(−6,8,−2). ‖−2u‖₁=6+8+2=16=2·8 and ‖−2u‖_∞=8=2·4. The factor is |−2|=2, not −2; a norm is never negative.
S15 · implies
Before
\varphi(v)=|v_1|\ \text{on}\ \mathbb{R}^3After
\varphi(0,1,0)=0,\ (0,1,0)\neq 0;\ \varphi\ \text{is not a norm}Conditions: none
Rule: quantifier_negation · Uses: G4, quantifier_negation, G3
Leaf basis: quantifier_negation + abs_rules + numeric_evaluation
φ satisfies nonnegativity, homogeneity (|a v₁|=|a||v₁|) and the triangle inequality (|u₁+v₁|≤|u₁|+|v₁|), all for every vector. But N1 requires φ(v)=0 only when v=0; v=(0,1,0)≠0 has φ(v)=|0|=0. One counterexample refutes this universal axiom, so φ is not a norm (it is only a seminorm).
S16 · implies
Before
\psi(v)=(\|v\|_1)^2,\ e_1=(1,0,0),\ e_2=(0,1,0)
After
\psi(e_1+e_2)=4>2=\psi(e_1)+\psi(e_2);\ \psi(2e_1)=4\neq 2=|2|\psi(e_1)
Conditions: none
Rule: quantifier_negation · Uses: G4, S06, quantifier_negation
Leaf basis: quantifier_negation + numeric_evaluation + power_laws
ψ≥0, and by S03 ψ(v)=0 ⇔ ‖v‖₁=0 ⇔ v=0, so N1 holds. But ψ(e₁+e₂)=(1+1)²=4 while ψ(e₁)+ψ(e₂)=1+1=2, so the triangle inequality fails; ψ(2e₁)=2²=4 while |2|ψ(e₁)=2, so homogeneity fails too. Hence ψ is not a norm.