EXACT CLAIM

Consider two independent uniform bits with qubit outputs ρ±^(1)=(I+aZ±√(1−a²)X)/2 and ρ±^(2)=(I+cZ±√y X)/2, where 0≤a,c≤1 and 0≤y≤1−c². Thus the first channel has pure conditional outputs. Entropies are in bits. Define h(t)=h₂((1+t)/2), J_c(y)=h(c)−h(√(c²+y)), and Δ=s₁+s₂−s₁s₂−H(X₁⊕X₂|B₁B₂). For a,c<1, put A=1−a², C=1−c², q=y/C, f(t)=atanh(t)/t with f(0)=1, L(t)=f(t)/(2 ln 2), T(a)=Af(a)/h(a), and F(q)=Σ_{n≥1}q^(n−1)/[n(2n−1)]. Then
Δ ≥ h(a)yL(c)[T(a)−F(q)].
Consequently the erasure upper bound holds whenever F(q)≤T(a). In particular, uniformly for 63/64≤a<1, 0≤c<1, and 0≤y≤(1−c²)/2,
Δ ≥ h(a)yL(c)/120.
This strip is strict when y>0. If a=1 or c=1, Δ=0. Also T(a)→2 ln 2 as a→1; hence for each fixed q<1 the sufficient condition holds for all a sufficiently close to 1, uniformly in c. This is a partial one-pure qubit result, not a resolution of the arbitrary-output conjecture.

FULL FROZEN PROOF

Write ℓ=ln 2. First establish the two analytic estimates used in the claim.

1. Parity information. For temporarily arbitrary equal-purity qubit signals b,d, let z=b²d² and let D=((I+aZ)/2)⊗((I+cZ)/2). The parity-conditioned states are D±√z X⊗X/4. They are unitarily equivalent and their average is D. Set K(z)=S(D)−S(D+√z X⊗X/4). Thus H(X₁⊕X₂|B₁B₂)=1−K(z).

The matrix splits into two 2-by-2 blocks, each with fixed trace w and eigenvalues (w±√((p−r)²+z/4))/2, where p,r are its diagonal entries. Its entropy is w h(√((p−r)²+z/4)/w)−w log₂w. The expansion
1−h(t)=Σ_{n≥1}t^{2n}/[2n(2n−1)ℓ]
shows that each block's entropy loss is convex as a function of z. This remains true at the physical boundary by continuity. Consequently K(z)≥zK′(0).

For a,c<1, direct differentiation of these blocks gives
K′(0)=[(a f(a)·a−c f(c)·c)/(a²−c²)]/(2ℓ),
with coincident arguments interpreted continuously. Equivalently,
K′(0)=(1/(2ℓ))∫₀¹ dt/[(1−a²t²)(1−c²t²)].
The two factors in this integrand are increasing functions of t. Chebyshev's integral inequality therefore gives
K′(0)≥f(a)f(c)/(2ℓ).
For completeness, Chebyshev here follows by integrating the nonnegative expression (u(t)−u(s))(v(t)−v(s)) over the unit square. We have proved K(z)≥z f(a)f(c)/(2ℓ).

2. Individual information. Integrating the derivative of J_c yields the nonnegative expansion
J_c(y)=Σ_{n≥1} yⁿ/(2nℓ) ∫₀¹ t^{2n−2}/(1−c²t²)ⁿ dt.
For y<C this follows by a geometric-series expansion, and at y=C it follows by monotone convergence. Substitute v=√C t/√(1−c²t²). The integral becomes
C^{1−n}∫₀¹ v^{2n−2}/√(C+c²v²) dv.
Here v^{2n−2} is increasing and the remaining factor is decreasing. Reverse Chebyshev, proved by the same double-integral identity with the opposite sign, bounds this by
C^{1−n}/(2n−1) ∫₀¹ dv/√(C+c²v²)
= C^{1−n}f(c)/(2n−1).
The formula includes c=0 by continuity. Summing proves
J_c(y)≤yL(c)F(y/C).

3. The one-pure gap. In the claimed family b²=A and d²=y. The individual Holevo informations are 1−s₁=h(a) and 1−s₂=J_c(y). Therefore exactly
Δ=K(Ay)−h(a)J_c(y).
Applying estimates 1 and 2 gives
Δ≥Ay f(a)f(c)/(2ℓ)−h(a)yL(c)F(q)
=h(a)yL(c)[T(a)−F(q)],
which proves the general sufficient condition, including y=0.

4. An explicit uniform strip. We now prove T(a)>9/8 for 63/64≤a<1 using elementary certified constants. Let δ=1−a, p=δ/2, and R=ln(2/δ). Natural binary entropy satisfies
−p ln p−(1−p)ln(1−p)≤p(R+1),
because −(1−p)ln(1−p)≤p. Consequently
T(a)≥ℓ(1+a)/a · ln((1+a)/δ)/(R+1)
≥2ℓ[R+ln(1−δ/2)]/(R+1).
Since δ≤1/64, ln(1−δ/2)≥−(δ/2)/(1−δ/2)≥−1/127>−1/100. Also
ℓ=2Σ_{k≥0}(1/3)^{2k+1}/(2k+1)>56/81>69/100,
so R≥ln128=7ℓ>483/100. The function (R−1/100)/(R+1) is increasing. These inequalities yield
T(a)≥(69/50)(482/583)=16629/14575>9/8.
All quantities multiplied in this argument are positive.

For 0≤q≤1/2, separating the first two terms of F and using n(2n−1)≥15 for n≥3 gives
F(q)≤1+q/6+q²/[15(1−q)]≤1+1/12+1/30=67/60.
Hence T(a)−F(q)≥9/8−67/60=1/120. Inserting this in step 3 proves the quantitative strip. For a<1, c<1, and y>0, its right-hand side is positive.

5. Endpoints and scope. If a=1, the first signal vanishes and s₁=1; the parity bit is independent of both outputs, so Δ=0. If c=1 then y=0, with the same conclusion because s₂=1. Finally, as δ=1−a tends to zero, natural binary entropy at δ/2 is (δ/2)[ln(2/δ)+1+o(1)], while Af(a)=δ(1+o(1))ln(2/δ). Thus T(a)→2ℓ. The entropy series gives F(1)=2ℓ, and its positive coefficients imply F(q)<F(1) for q<1. The asserted sufficiently-large-a conclusion follows directly from this limit, without assuming monotonicity of T.

The accompanying exact-rational checker verifies the decisive constants in step 4; the entropy inequalities themselves have been proved analytically above.

REMAINING GAP

The condition F(q)≤T(a) does not settle the complete one-pure boundary, particularly strong second signals at general a. The full equal-purity qubit family and the erasure upper bound for arbitrary quantum side information remain unresolved.