Main derivation
S01 · definition
Before
B\ \text{bilinear}After
\forall y:\ x\mapsto B(x,y)\ \text{linear}\ \land\ \forall x:\ y\mapsto B(x,y)\ \text{linear}Conditions: none
Rule: mko_bilinear_def · Uses: G1, linear_map_def
Leaf basis: mko_bilinear_def + def_unfold
By the MKO, bilinearity has two separate obligations: with the second input fixed the first slot is linear, and with the first input fixed the second slot is linear. Both must hold for all vectors and all real coefficients.
S02 · equivalent
Before
x^{T}Ay\ \ (1\times2)(2\times2)(2\times1)After
B(x,y)=\sum_{i=1}^{2}\sum_{j=1}^{2}x_i\,a_{ij}\,y_jConditions: \text{dimensions compatible}
Rule: matrix_product_def · Uses: G1, matrix_product_def
Leaf basis: matrix_product_def + distributive + finite_sum_rules
Dimensions are compatible: x^T is 1×2, A is 2×2, y is 2×1, so the product is 1×1, a real number. Apply the definition of the matrix product twice: (Ay)_i=Σ_j a_ij y_j, then x^T(Ay)=Σ_i x_i (Ay)_i, which together give a double sum.
S03 · equivalent
Before
\sum_{i,j}x_i\,a_{ij}\,y_jAfter
x_1y_1+2x_1y_2-x_2y_1+3x_2y_2
Conditions: none
Rule: finite_sum_rules · Uses: S02, G1
Leaf basis: finite_sum_rules + numeric_evaluation
Write out the four terms (i,j)=(1,1), (1,2), (2,1), (2,2) and substitute a11=1, a12=2, a21=−1, a22=3. N1 checks this multilinear identity on a grid.
S04 · equivalent
Before
B(\alpha x+\beta x',y)
After
\sum_{i,j}(\alpha x_i+\beta x'_i)\,a_{ij}\,y_jConditions: none
Rule: vector_space_axioms · Uses: S02, vector_space_axioms
Leaf basis: vector_space_axioms + eq_replace
First slot: fix y and take any x, x′∈ℝ² and α, β∈ℝ. The i-th coordinate of αx+βx′ is αx_i+βx′_i (coordinate operations); substitute into the double sum of S02.
S05 · equivalent
Before
\sum_{i,j}(\alpha x_i+\beta x'_i)\,a_{ij}\,y_jAfter
\sum_{i,j}\left(\alpha\,x_i a_{ij} y_j+\beta\,x'_i a_{ij} y_j\right)Conditions: none
Rule: distributive · Uses: S04, distributive
Leaf basis: distributive + mul_assoc + mul_comm
Expand each term by distributivity: (αx_i+βx′_i)a_ij y_j=αx_i a_ij y_j+βx′_i a_ij y_j.
S06 · equivalent
Before
\sum_{i,j}\left(\alpha\,x_i a_{ij} y_j+\beta\,x'_i a_{ij} y_j\right)After
\alpha B(x,y)+\beta B(x',y)
Conditions: none
Rule: finite_sum_rules · Uses: S05, finite_sum_rules
Leaf basis: finite_sum_rules + eq_replace
A finite sum splits into two sums (Σ(a+b)=Σa+Σb), and the constants α, β come out (Σca=cΣa); by S02 the remaining sums are B(x,y) and B(x′,y). So the first slot is linear. N2 checks this identity on a grid.
S07 · theorem_application
Before
B(x,\alpha y+\beta y')
After
\alpha B(x,y)+\beta B(x,y')
Conditions: none
Rule: L_slot2 · Uses: S02
Leaf basis: L_slot2
The second slot must be proved separately (Lemma L_slot2, S07.1 to S07.3): A is not symmetric, so the first-slot result cannot be carried over by 'swapping x and y'.
S08 · implies
Before
\text{slot 1 linear (S06)},\ \text{slot 2 linear (S07)}After
B\ \text{is a bilinear form on}\ \mathbb{R}^2Conditions: none
Rule: forall_intro · Uses: S01, S06, S07
Leaf basis: forall_intro + and_intro_elim + mko_bilinear_def
In S04 to S07, x, x′, y, y′, α, β were arbitrary with no extra assumption, so by generalization both slots are linear for all inputs, and B is bilinear. The argument used only that the a_ij are fixed numbers, so a fixed m-by-n real matrix gives a bilinear map on the compatible spaces ℝᵐ×ℝⁿ. A bilinear form on one ℝⁿ uses an n-by-n matrix; this example is the square n=2 case.
S09 · implies
Before
B(x,y)=x^{T}(Ay),\ x=(2,-1),\ y=(1,4)After
B(x,y)=7
Conditions: none
Rule: matrix_product_def · Uses: S02, G1, G2
Leaf basis: matrix_product_def + numeric_evaluation
First compute Ay (S09.1), then multiply by x^T (S09.2).
S10 · implies
Before
B(y,x)=y^{T}(Ax),\ Ax=(1\cdot2+2\cdot(-1),\ -1\cdot2+3\cdot(-1))After
Ax=(0,-5),\ B(y,x)=1\cdot0+4\cdot(-5)=-20
Conditions: none
Rule: matrix_product_def · Uses: S02, G1, G2
Leaf basis: matrix_product_def + numeric_evaluation
Swap the two inputs: Ax=(2−2,−2−3)=(0,−5) and y^T(Ax)=1·0+4·(−5)=−20.
S11 · implies
Before
B(x,y)=7,\ B(y,x)=-20
After
B\ \text{bilinear but not symmetric}Conditions: none
Rule: numeric_evaluation · Uses: S08, S09, S10
Leaf basis: numeric_evaluation + quantifier_negation
7≠−20, so B is not symmetric; yet S08 proves B bilinear. Symmetry is an extra condition, not part of bilinearity; an inner product additionally needs symmetry and positive definiteness, so B is not an inner product.
S12 · implies
Before
\alpha x+\beta x'=(4,-5)
After
B((4,-5),y)=-19=3\cdot7+(-2)\cdot20
Conditions: none
Rule: numeric_evaluation · Uses: S06, S09, G2
Leaf basis: numeric_evaluation + matrix_product_def
αx+βx′=(3·2−2·1, 3·(−1)−2·1)=(4,−5). With Ay=(9,11): B((4,−5),y)=36−55=−19; B(x′,y)=9+11=20, so αB(x,y)+βB(x′,y)=21−40=−19. This instance agrees with S06 but is an example, not a proof.
S13 · implies
Before
x=x'=(0,1),\ y=(0,1)
After
f(x+x',y)=6\neq0=f(x,y)+f(x',y)
Conditions: none
Rule: contradiction · Uses: G3, linear_map_def
Leaf basis: contradiction + quantifier_negation + numeric_evaluation
f((0,2),(0,1))=0+(8−2)·1=6, while f((0,1),(0,1))=0+(1−1)·1=0, and the sum of two such values is 0. With α=β=1 linearity in the first slot fails, so f is not bilinear.
S14 · implies
Before
x_2,\ x'_2,\ x_2+x'_2\in\{-1,0,1\}After
f(x+x',y)=f(x,y)+f(x',y)\ \text{on these samples}Conditions: none
Rule: numeric_evaluation · Uses: G3, S13
Leaf basis: numeric_evaluation + power_laws
When t∈{−1,0,1}, t³−t=0, so whenever x2, x′2 and x2+x′2 all lie in {−1,0,1}, f equals x1y1 on those inputs and every additivity sample passes (for example x=(1,1), x′=(2,−1)). Yet S13 shows f is not bilinear: passing finitely many samples cannot prove a black-box function bilinear; a universal certificate like S04 to S07 is needed.
Substeps and lemmas
S07.1 · equivalent
Before
B(x,\alpha y+\beta y')
After
\sum_{i,j}x_i\,a_{ij}\,(\alpha y_j+\beta y'_j)Conditions: none
Rule: vector_space_axioms · Uses: S02, vector_space_axioms
Leaf basis: vector_space_axioms + eq_replace
Fix x; the j-th coordinate of αy+βy′ is αy_j+βy′_j, substituted into the double sum.
S07.2 · equivalent
Before
\sum_{i,j}x_i\,a_{ij}\,(\alpha y_j+\beta y'_j)After
\sum_{i,j}\left(\alpha\,x_i a_{ij} y_j+\beta\,x_i a_{ij} y'_j\right)Conditions: none
Rule: distributive · Uses: S07.1, distributive
Leaf basis: distributive + mul_comm + mul_assoc
Expand each term by distributivity and move α, β to the front (commutativity and associativity of multiplication).
S07.3 · equivalent
Before
\sum_{i,j}\left(\alpha\,x_i a_{ij} y_j+\beta\,x_i a_{ij} y'_j\right)After
\alpha B(x,y)+\beta B(x,y')
Conditions: none
Rule: finite_sum_rules · Uses: S07.2, finite_sum_rules
Leaf basis: finite_sum_rules + eq_replace
Split into two sums and pull out the constants; by S02 this is αB(x,y)+βB(x,y′). S07.1 to S07.3 together prove Lemma L_slot2. N3 checks it on a grid.
S09.1 · equivalent
Before
Ay=\begin{pmatrix}1&2\\-1&3\end{pmatrix}\begin{pmatrix}1\\4\end{pmatrix}After
\begin{pmatrix}9\\11\end{pmatrix}Conditions: none
Rule: matrix_product_def · Uses: G1, G2
Leaf basis: matrix_product_def + numeric_evaluation
(Ay)_1=1·1+2·4=9 and (Ay)_2=(−1)·1+3·4=11.
S09.2 · equivalent
Before
x^{T}(Ay)=(2,-1)\begin{pmatrix}9\\11\end{pmatrix}After
2\cdot9+(-1)\cdot11=7
Conditions: none
Rule: matrix_product_def · Uses: S09.1, G2
Leaf basis: matrix_product_def + numeric_evaluation
2·9+(−1)·11=18−11=7.
Error controls and negative controls
Symmetry is not bilinearity
The mistake named by the MKO is to declare bilinearity or an inner product after checking only B(x,y)=B(y,x). g(x,y)=(x1y1)² is symmetric, but g(2e1,e1)=4≠2=2g(e1,e1), so it is not bilinear. Conversely B here is bilinear but not symmetric (7≠−20). B(x,x)=x1²+x1x2+3x2²=(x1+x2/2)²+(11/4)x2² is positive for x≠0, yet B is not an inner product because it is not symmetric.
Check each slot separately; bilinear is not linear in the pair
A is not symmetric, so the second slot cannot be obtained from the first by swapping (S07 proves it separately). Bilinear also does not mean B(x+x′,y+y′)=B(x,y)+B(x′,y′): the correct expansion has the cross terms B(x,y′)+B(x′,y). For example x=x′=y=y′=e1: B(2e1,2e1)=4 while B(e1,e1)+B(e1,e1)=2.
Black-box samples are not enough
f passes additivity on every sample with x2, x′2, x2+x′2∈{−1,0,1} (S14), yet fails at (0,1)+(0,1) (S13). MKO assumption a2: matrix expansion is a universal certificate, and finite samples cannot prove an arbitrary black-box function bilinear.
- g(x,y)=(x1y1)² is symmetric, so it is a bilinear form →
reject — g(2e1,e1)=4 while 2g(e1,e1)=2. - Since B(x,y)=7≠−20=B(y,x), x^T A y is not a bilinear form →
reject — Bilinearity does not require symmetry; S04 to S08 prove B bilinear. - Bilinear means B(x+x′,y+y′)=B(x,y)+B(x′,y′) →
reject — With x=x′=y=y′=e1 the left side is 4 and the right side 2; the cross terms are missing. - f passed several additivity tests such as x=(1,1), x′=(2,−1), so f is bilinear →
reject — f((0,2),(0,1))=6 while f((0,1),(0,1))+f((0,1),(0,1))=0.