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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Self-supervised and contrastive representations — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A negative sample collides with the positive class

Example

Let the latent classes be a,b with equal probabilities, let each class contain one deterministic input, and define a scalar representation by f(a)=0 and f(b)=1. Use one independent marginal negative and the hinge loss. The probability that the negative class equals the positive-pair class is τ=1/2. Conditional on such a collision, the contrastive score is always zero and the conditional loss is 1. Conditional on distinct classes, the ordered pair (a,b) has loss 1 and (b,a) has loss 0, so Lun≠(f)=1/2 and the full contrastive risk is Lun(f)=3/4.

Example references

Latent-class contrastive representation learning: defines the positive-pair and independent-negative sampling laws.

Convexity, Lipschitz bounds, and oscillation of the source contrastive losses: defines the hinge loss used in the calculation.

Saunshi et al., Section 4.2 and Equation (8): decomposes the objective by whether the negative class collides with the positive class.

Facts & Assumptions

Given: Two equally likely classes a,b, deterministic inputs, f(a)=0, f(b)=1, one independent marginal negative, and hinge loss.

[F1]

The positive and negative classes are independent with law (1/2,1/2), so equality has probability 1/2 and inequality has probability 1/2. Latent-class contrastive representation learning

[F2]

The hinge contrastive loss is (1+t)+ and equals 1 at score t=0. Convexity, Lipschitz bounds, and oscillation of the source contrastive losses

Proof

Given: The finite sampling model in the Example.

1.1F1F2givenalgebra

If the negative class equals the positive class, the anchor, positive sample and negative sample all have the same representation. Thus qf=f(x)(f(x−)−f(x+))=0 and [F2] gives collision loss 1.

1.2F1F2givenalgebra

If the ordered classes are (a,b), the anchor representation is 0, hence qf=0 and loss 1. If they are (b,a), then qf=1(0−1)=−1, hence the hinge loss is 0. The two ordered cases are equiprobable given inequality, so Lun≠(f)=(1+0)/2=1/2.

2.1F1step 1.1step 1.2givenalgebra∎

The collision decomposition weights the two conditional losses by τ=1/2 and 1−τ=1/2. Therefore Lun(f)=(1/2)⋅1+(1/2)⋅(1/2)=3/4, with a positive contribution from negative-class collisions.

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

High within-class variance raises contrastive loss despite perfect mean separation

Example

Let C={a,b} be uniform and let each class contain two equally likely inputs indexed by η∈{−1,+1}. In R2, define f(a,η)=(1,ηr) and f(b,η)=(−1,ηr) for r≥1. Each class mean is respectively (1,0) and (−1,0), so the mean classifier separates both classes with hinge loss zero. Yet for one independent marginal negative and hinge contrastive loss, the total risk is Lun(f)=14+12r2, which grows without bound as the within-class variance r2 increases. The norm bound holds with R=1+r2 for each fixed r.

Example references

Latent-class contrastive representation learning: defines the latent-class triplet law, mean classifier, covariance and contrastive risk.

Convexity, Lipschitz bounds, and oscillation of the source contrastive losses: defines the hinge loss.

Saunshi et al., Section 5.1 and Figure 1(b): gives a two-class example where high within-class variation raises the unsupervised loss despite a good mean classifier.

Facts & Assumptions

Given: The two-class distribution and representation in the Example, one independent marginal negative, and hinge loss.

[F1]

The class means are μa=(1,0) and μb=(−1,0); the mean classifier for the ordered task (a,b) has weight (2,0). Latent-class contrastive representation learning

[F2]

The class-mean hinge risk uses loss (1−Yw⊤f(X))+, and the contrastive hinge loss is (1+qf)+. Latent-class contrastive representation learning, Convexity, Lipschitz bounds, and oscillation of the source contrastive losses

[F3]

Given the positive class, the negative class equals it with probability 1/2; conditional on collision, the score's random part is r2η0(η−−η+), whose values 2r2,−2r2,0 have probabilities 1/4,1/4,1/2. Conditional on distinct classes, the score is −2+r2η0(η−−η+) with the same three probabilities. Latent-class contrastive representation learning

Proof

Given: The finite class-conditional laws in the Example.

1.1F1F2givenalgebra

For the task (a,b), the mean weight is (2,0). On class a with label +1, Yw⊤f(a,η)=2 for either η; on class b with label −1, it is also 2. The hinge loss is therefore zero for both labels. Swapping the ordered task gives the same result, so the average mean-classifier hinge risk is zero.

1.2F2F3givenalgebra

Conditional on a class collision, [F3] gives score values 2r2,−2r2,0 with probabilities 1/4,1/4,1/2. Since r≥1, their hinge losses are 1+2r2,0,1, respectively. Thus Lun=(f)=14(1+2r2)+12=34+12r2.

1.3F2F3givenalgebra

Conditional on distinct classes, the score values are −2+2r2,−2−2r2,−2 with probabilities 1/4,1/4,1/2. For r≥1, their hinge losses are 2r2−1,0,0. Hence Lun≠(f)=14(2r2−1)=12r2−14.

2.1F3step 1.2step 1.3step 1.1givenalgebra∎

Collision and distinct-class events each have probability 1/2. Combining steps 1.2 and 1.3 yields Lun(f)=12(34+12r2)+12(12r2−14)=14+12r2. This tends to infinity with r, while step 1.1 keeps the mean-classifier hinge risk zero.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

An augmentation that erases the label-relevant class coordinate

Example

Let the raw input be (c,n)∈{−1,+1}2, where the latent class is c and, conditional on either class, the nuisance coordinate n is uniform on {−1,+1}. Apply the augmentation T(c,n)=n before forming positive pairs and negative samples. The resulting input law Dc is the same uniform law for both classes, so Dsim=Dneg⊗Dneg and the observed triplets carry no class-dependent distributional signal. Consequently, every deterministic representation of the augmented input is independent of the binary task label, and its optimal average hinge risk for a uniform-label two-class task is exactly 1; the erased class coordinate is unavailable to both the contrastive learner and the downstream linear classifier.

Example references

Latent-class contrastive representation learning: defines the class-conditional post-augmentation laws and binary tasks.

Convexity, Lipschitz bounds, and oscillation of the source contrastive losses: defines the hinge loss used to compute the downstream risk.

Chen et al., Section 3: explains how augmentation choices define the predictive tasks in contrastive learning.

Facts & Assumptions

Given: The two latent classes, the nuisance law, the deterministic augmentation T(c,n)=n, and a deterministic representation f of the augmented input.

[F1]

After augmentation, D+1=D−1 is uniform on {−1,+1}, so the observed input has the same law under both labels. Latent-class contrastive representation learning

[F2]

For a fixed score z=w⊤f(x), the average hinge loss over uniform labels is 12[(1−z)++(1+z)+], which is at least 1 for every real z; choosing w=0 gives risk 1. Convexity, Lipschitz bounds, and oscillation of the source contrastive losses

Proof

Given: The finite model in the Example.

1.1F1given

Since T(c,n)=n and n is uniform independently of c, each post-augmentation class law is the same uniform law U. The mixture formula gives Dneg=U and Dsim(A×B)=U(A)U(B) for all subsets A,B of the finite input space, hence Dsim=U⊗U. Thus the observed triplet law is U⊗U⊗U, independent of the latent task labels, and every score z=w⊤f(n) has the same distribution under both labels.

2.1F1F2step 1.1givenalgebra∎

For each fixed score z, if ∣z∣≤1 then (1−z)++(1+z)+=2. If z≥1, the sum is 0+(1+z)≥2; if z≤−1, it is (1−z)+0≥2. Therefore [F2] gives average risk at least 1 after integrating over the common input law, and w=0 attains 1. The class coordinate erased by T cannot be recovered by this downstream classifier.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

SimCLR reports an empirical projection-head effect

Example

In SimCLR's ImageNet experiments, a projection head is a map g applied to the encoder representation h to produce z=g(h); linear evaluation trains a linear classifier on a fixed representation. The authors report a comparison among an identity head, a linear projection head, and a nonlinear projection head with one additional hidden layer and ReLU. In their Figure 8, the nonlinear head's pre-projection representation h scores about three percentage points above the linear-head condition and more than ten points above the no-head condition; they also report that h scores more than ten points above the projected representation z=g(h). Their default setup uses a ResNet-50 encoder, a two-layer MLP projection head to a 128-dimensional latent space, NT-Xent (normalized temperature-scaled cross-entropy), batch size 4096 and 100 training epochs. These are reported empirical results under the paper's evaluation protocol, not a theorem that projection heads universally improve representations.

Example references

Latent-class contrastive representation learning: supplies the representation-learning context only; the reported benchmark is empirical.

Chen et al., Section 4 default setup and Section 4.2, Figure 8: reports the encoder/head configuration and linear-evaluation comparisons.

Chen et al., Supplementary Appendix B.4: provides additional projection-head analysis.

Facts & Assumptions

Given: The published SimCLR paper and its ImageNet projection-head experiments.

[F1]

The default configuration in Section 4 uses a ResNet-50 base encoder, a two-layer MLP projection head to dimension 128, NT-Xent, batch size 4096 and 100 epochs. Chen et al., Section 4, default setting

[F2]

Figure 8 reports linear evaluation for identity, linear and nonlinear projection heads; its text reports roughly a three-point gain over a linear projection, more than ten points over no projection, and more than ten points for pre-projection h over z=g(h). Chen et al., Section 4.2 and Figure 8

Proof

Given: The complete relevant passages of the SimCLR paper inspected in [F1] and [F2].

1.1F1given

The setup in [F1] identifies the encoder, projection head, latent dimension, loss, batch size and training duration for the reported default experiment.

2.1F2step 1.1given∎

The linear-evaluation comparisons in [F2] state the reported relative performance of the three head choices and of h versus g(h). The Example reports those empirical comparisons with their stated experimental scope and makes no universal claim.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

The hinge margin comparison does not extend to logistic contrastive loss

Example

Consider two equally likely deterministic classes a,b and the scalar representation f(a)=2, f(b)=4, so R=4, τ=1/2, and the classwise centered representation has sub-Gaussian proxy σ2=0. For the distinct-class tasks, the mean classifier has logistic risk

Lsupμ(f)=12[log⁡2(1+e4)+log⁡2(1+e−8)]>52,

while its source hinge margin risk at γ=1 is exactly L1,supμ(f)=5/2. The logistic contrastive risk conditioned on distinct classes has the same value as the displayed logistic mean-classifier risk. Consequently, for every 0<ϵ≤1 smaller than the positive gap, the source hinge margin comparison Lun≠(f)≤γLγ,supμ(f)+ϵ fails for logistic contrastive loss. In particular, the hinge-margin corollary cannot be transferred unchanged to logistic training.

For the optional transfer comparison, assume the Axiom of Choice, fix M≥1 and 0<δ<1, impose the corollary's remaining model, risk and supremum measurability, and measurable exact-empirical-minimizer hypotheses, and use the singleton F={f}. The proposed extension is

Lsupμ(f^)≤γLγ,supμ(f)+ϵ+2L0τs(f)+2ΓMlog(S,δ)1−τ,

where Lsupμ(f^) uses logistic task loss, Lγ,supμ(f) uses the scaled hinge margin loss, γ=1 here, and

ΓMlog(S,δ):=218L0R RS(F)M+3Bℓlog,Rln⁡(4/δ)2M

uses the logistic loss oscillation. The example tests this claimed mean-classifier bound with logistic contrastive training.

Example references

Latent-class contrastive representation learning: defines the latent-class risks, scores and centered classwise representations.

Convexity, Lipschitz bounds, and oscillation of the source contrastive losses: defines the base-2 logistic loss and hinge margin loss.

Class-conditional sub-Gaussian concentration yields a contrastive margin bound: states the one-negative margin inequality for hinge contrastive loss.

Sub-Gaussian contrastive transfer with a mean-classifier margin loss: states the source margin transfer corollary with its hinge-only scope.

The Axiom of Choice: supplies the choice assumption for the optional transfer comparison.

Saunshi et al., Lemma 5.1 and Corollary 5.1.1: states the margin comparison in the hinge-loss section.

Facts & Assumptions

Given: The deterministic two-class model and representation in the Example; base-2 logistic contrastive loss; hinge loss for the scaled mean-classifier margin; and 0<ϵ≤1. For the optional transfer comparison, also assume AC, M≥1, 0<δ<1, the stated risk and supremum measurability hypotheses, and a measurable exact empirical selector for F={f}.

[A1]

The optional transfer comparison assumes the Axiom of Choice required by the transfer corollary's uniform-generalization route. The Axiom of Choice, Sub-Gaussian contrastive transfer with a mean-classifier margin loss

[F1]

The ordered distinct class pairs (a,b) and (b,a) have equal conditional probability, the means equal their deterministic embeddings, and the positive-pair and negative sampling rules are those in the latent-class model. Latent-class contrastive representation learning

[F2]

The contrastive score is qf=f(x)⊤(f(x−)−f(x+)), logistic loss is log⁡2(1+eq), and the scaled hinge mean-classifier loss is (1−Ywμ⊤f(x)/γ)+. Latent-class contrastive representation learning, Convexity, Lipschitz bounds, and oscillation of the source contrastive losses

[F3]

The source sub-Gaussian margin parameter equals γ=1 when σ=0, and the source margin corollary is hinge-only. Class-conditional sub-Gaussian concentration yields a contrastive margin bound, Sub-Gaussian contrastive transfer with a mean-classifier margin loss

Proof

Given: The finite deterministic model in the Example.

1.1F1F2givenalgebra

The class means are μa=2 and μb=4. For task (a,b), the mean weight is wμ=μa−μb=−2. The positive-label example a has logistic argument −wμf(a)=4, while the negative-label example b has argument wμf(b)=−8. Swapping the classes gives the same two losses in reverse order, proving the displayed mean-classifier logistic risk.

1.2F2F3givenalgebra

For the mean classifier, the two hinge-margin arguments at γ=1 are 4 and −8, so the losses are 5 and 0 and their average is 5/2. The sub-Gaussian proxy is zero because each class is deterministic, so [F3] gives γ=1 for every permitted ϵ.

2.1F1F2step 1.1givenalgebra

For distinct classes, the contrastive score is qf=2(4−2)=4 for (a,b) and qf=4(2−4)=−8 for (b,a). Therefore Lun≠(f)=12[log⁡2(1+e4)+log⁡2(1+e−8)], equal to the value in step 1.1.

3.1F2step 2.1step 1.2givenalgebra

The exponential series has positive terms and e4>1+4+42/2!+43/3!+44/4!=103/3>31. Hence 1+e4>32, so log⁡2(1+e4)>5; also log⁡2(1+e−8)>0. Thus the value in step 2.1 exceeds 5/2 by a positive gap Δ. Choose 0<ϵ<min⁡(Δ/4,1). Then Lun≠(f)>L1,supμ(f)+ϵ, contradicting the proposed logistic extension.

4.1A1F1F2F3step 1.2givenalgebra∎

Under [A1] and the transfer hypotheses in the Statement, take F={f} and the logistic objective. The only element is an exact empirical minimizer, its vector Rademacher complexity is zero because the expectation of its finite signed sum is zero, and s(f)=0 because both class covariances vanish. Therefore ΓMlog(S,δ)=3Bℓlog,4ln⁡(4/δ)/(2M)→0 as M→∞ for the fixed 0<δ<1. The mean-classifier logistic risk Lsupμ(f) exceeds 5/2 by Δ, while the proposed logistic-training extension would upper-bound that risk by 5/2+ϵ+4ΓMlog(S,δ) (since 1−τ=1/2). Taking 0<ϵ<min⁡(Δ/4,1) and then M large enough that 4ΓMlog<Δ/4 makes that right side less than 5/2+Δ/2, contradicting Lsupμ(f)=5/2+Δ.

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