EXACT CLAIM

A constructive exclusion for an explicit non-transpose-symmetric PPT-entangled family. Let A=E=C³, let 0<t<1 and s=1−t², and use the computational product basis |ij>. Define p=|00>+|11>+|22>, v=|20>+t|22>, and H_t=s[|p><p|+∑_{ij∈{01,02,10,12,21}}|ij><ij|]+|v><v|. Put M_t=diag(3s,3s,s+2), W_t=M_t^(−1/2), and J_t=(W_t⊗I)H_t(W_t⊗I). In the convention Tr_E J(Ψ)=I_A, let Ψ_t:L(C³)→L(C³) have Choi operator J_t. Then J_t is PPT entangled, rank(J_t)=7 and rank(J_t^{T_E})=6. Nevertheless Ψ_t is antidegradable, and every minimal complementary channel Φ_t:L(C³)→L(C⁷) is ordinarily degradable. Thus this entire family cannot furnish a transpose-degradable but nondegradable channel by taking Φ_t. The rank mismatch also rules out J_t^{T_E}=(I⊗U)J_t(I⊗U†) for any output unitary U. No transpose-degradability assertion about Φ_t is required or made.

FULL FROZEN PROOF

Write Γ for partial transpose on E. All vectors and coefficients defining H_t are real. The seven vectors p,v,|01>,|02>,|10>,|12>,|21> are linearly independent, and their coefficients in H_t are strictly positive. Consequently H_t≥0 and rank(H_t)=7. Direct partial trace gives Tr_E H_t=diag(3s,3s,s+2)=M_t. Since s>0, W_t exists and J_t is a positive operator with input marginal I_A, hence is the Choi operator of a channel.

PPT and ranks. The partial transpose H_t^Γ is block diagonal in the following orthogonal decomposition. On |00> and |11> its singleton blocks are s. On each ordered pair (|01>,|10>) and (|12>,|21>) its block is s[[1,1],[1,1]]. On (|02>,|20>,|22>) its block is
  [[s,s,0],[s,1,t],[0,t,1]] = s f f^T + g g^T,
where f=(1,1,0)^T and g=(0,t,1)^T. This last identity uses s+t²=1. The last block is positive of rank two because f and g are independent; the other nonsingleton blocks each have rank one. Hence H_t^Γ≥0 and rank(H_t^Γ)=2+1+1+2=6. Invertible congruence by W_t⊗I preserves both ranks and positivity, and commutes with Γ. These assertions therefore hold for J_t. Unitary conjugation preserves rank, proving the stated absence of output-unitary transpose symmetry.

Entanglement. Two vectors in ker H_t are |00>−|11> and |00>+t|20>−|22>. Three vectors in ker H_t^Γ are |01>−|10>, |12>−|21>, and |02>−|20>+t|22>. Suppose nonzero x,y satisfy x⊗y∈ran H_t and x⊗conj(y)∈ran H_t^Γ. Orthogonality to these real kernel vectors gives
  x0 y0=x1 y1,
  x0 y0+t x2 y0−x2 y2=0,
  x0 conj(y1)=x1 conj(y0),
  x1 conj(y2)=x2 conj(y1),
  x0 conj(y2)−x2 conj(y0)+t x2 conj(y2)=0.
If y1≠0, the third and fourth equations imply x=c conj(y), where c=x1/conj(y1). Here c≠0, since otherwise x=0. Substitution into the fifth equation gives tc conj(y2)²=0, so y2=x2=0. The second equation now gives x0 y0=0, and the first then gives x1 y1=0, a contradiction. Thus y1=0. Since y≠0, the third and fourth equations now force x1=0.

If H_t were separable, it would be a finite positive sum of product projectors |x⊗y><x⊗y|. Each summand vector belongs to ran H_t: a vector in the kernel of a positive sum must be orthogonal to each summand. Applying the same observation to the partially transposed decomposition puts x⊗conj(y) in ran H_t^Γ. The preceding argument would therefore make every summand have zero |11> component. This contradicts <11|H_t|11>=s>0. Hence H_t is entangled. Invertible local filtering preserves separability in both directions, so J_t is entangled as well.

Positive symmetric extension. Introduce E'≅C³ and order tensor factors A,E,E'. Set
  w=|000>+|101>+|110>+|202>+|220>+t|222>,
  z=|2>⊗(|0>+t|2>)⊗(|0>+t|2>),
  Ω_t=s[|w><w|+|011><011|+|022><022|+|122><122|+|211><211|]+[t²/(1+t²)]|z><z|.
Every coefficient is positive, so Ω_t≥0. Each displayed vector is invariant under exchanging E and E', hence Ω_t has equal marginals on AE and AE'. To calculate the AE marginal, slice w according to the E' index. Its three slices are p, |10>, and v. Thus Tr_{E'}|w><w|=|p><p|+|10><10|+|v><v|. The four remaining basis projectors contribute |01>,|02>,|12>,|21>, respectively. Finally, Tr_{E'}|z><z|=(1+t²)|v><v|. It follows that Tr_{E'}Ω_t equals H_t, since the coefficient of |v><v| is s+t²=1. By exchange symmetry, the other marginal also equals H_t.

Let K_t=(W_t⊗I⊗I)Ω_t(W_t⊗I⊗I). Its two marginals are J_t and its input marginal is I_A. Hence K_t is the Choi operator of a channel Λ_t:A→E⊗E' whose two output marginal channels are both Ψ_t.

Degrading map and all kernel freedom. Choose any minimal Stinespring isometry U:A→E⊗B for Ψ_t, where dim B=rank J_t=7, and write Φ_t=Tr_E(U·U†). Choose a Stinespring isometry Q:A→E⊗E'⊗F for Λ_t. Viewed as a dilation of its first marginal Ψ_t, Q dilates the same channel as U. Minimal Stinespring uniqueness supplies an isometry R:B→E'⊗F such that Q=(I_E⊗R)U. Define D:L(B)→L(E') by D(Y)=Tr_F(RYR†). This is completely positive and trace preserving. Taking the second marginal of Λ_t gives D∘Φ_t=Ψ_t, identifying E' with E. Thus Φ_t is ordinarily degradable, equivalently Ψ_t is antidegradable. This construction establishes existence of a valid degrading channel without assuming uniqueness of a superoperator inverse or restricting its kernel freedom. The argument applies to every minimal choice of U.

REMAINING GAP

The original existence problem remains unresolved. This proves only that the specified family of complementary channels is ordinarily degradable; it neither supplies a strict example elsewhere nor proves general containment of transpose-degradable channels in degradable channels.