The universal equality in QIQC Eq.(3) is false. For d=2,n=4,delta=1/2,s=121/1280, the stabilizer branch E(X)=<0000|X|0000>|0><0| has fidelity547/726=3/4+5/1452. More strongly, a deterministic five-copy stabilizer protocol atdelta=2/3 has fidelity at least11693/17496=2/3+29/17496 for EVERY pure qubit target; its Haar average is325/486=2/3+1/486 and its success is identically1. The protocol randomly chooses a signed Pauli axis, measures four inputs in that basis, prepares the selected eigenstate if all outcomes match it, and otherwise retains the untouched fifth copy. Discarding extra copies and independent acceptance gives the same pointwise guarantee for all n>=5 and all prescribed0<s<=1. These are explicit feasible lower bounds, not claims of optimality. The four-copy protocol alone is not pointwise improvement.

Four-copy postselection and five-copy deterministic stabilizer purification

Let t=1-delta and E(X)=<0000|X|0000>|0><0|. This is implementable by computational-basis measurements of all four inputs, acceptance only of0000, and preparation of|0>. Its single Kraus operator is K=|0><0000|, K^dagger K=|0000><0000|<=I. It is completely stabilizer preserving as an actual stabilizer-measurement branch with stabilizer preparation, including arbitrary references; its Choi projector is the five-qubit all-zero stabilizer projector. No unknown state enters the protocol.

For pure qubit psi with Bloch z coordinate z uniform on[-1,1], success is((1+tz)/2)^4 and output target overlap is(1+z)/2. Exact integration yields
D=1+2t^2+t^4/5,
s=D/16,
N=(D+4t/3+4t^3/5)/32,
F=N/s=1/2+(4t/3+4t^3/5)/(2D),
F-(1+t)/2=t(5-18t^2-3t^4)/(30D).
At t=1/2, delta=1/2:
s=121/1280, N=547/7680, F=547/726,
F-3/4=5/1452>0,
N-(3/4)s=1/3072>0.

This appears to refute the literal universal equality for n=4. The operation does NOT improve each individual pure state: its gain exploits a changed conditional distribution of accepted inputs. It does NOT contradict the two-copy analytical no-go theorem. Check this distinction explicitly against source definitions, not informal use of "universal".

Threshold: positive exactly when0<t<sqrt((-9+4sqrt(6))/3), equivalently sufficiently high noise. Adding discarded inputs extends to every n>=4. Independent thinning extends to any smaller prescribed average s. Mixing E with the discard-three/retain-one deterministic map gives positive integrated gain for any s between s0 and1, excluding1 itself. Do not claim the globally optimal fidelity or deterministic no-go refutation from this construction.



## A deterministic five-copy Haar counterexample

Let rho_t(r)=(I+t r.sigma)/2, psi(r)=(I+r.sigma)/2, f_t=(1+t)/2. Measure the first four qubits in the Z basis. On outcome0000 output a freshly prepared|0>; on every other outcome output the unmeasured fifth input. This is a trace-preserving stabilizer instrument followed by classical selection and discarding. Conditional on the unknown pure target, the fifth input remains rho_t(r) independent of the four outcomes. Thus its unconditional output fidelity is

f_t+((1+t r_z)/2)^4*((1+r_z)/2-f_t).

Its Haar average is f_t+N4-f_t*s4. At t=1/2 the exact four-copy integrals s4=121/1280,N4=547/7680 give

F_det=3/4+1/3072=2305/3072>3/4.

No success-conditioning distinction remains: success is identically1 on every input. This does not imply improvement for every target, since the gain can be negative at some individual r.

## A deterministic five-copy pointwise counterexample

Choose uniformly one of the six single-qubit Pauli eigenstates|v>, whose unit Bloch vectors are v in{+/-e_x,+/-e_y,+/-e_z}. Measure each of the first four inputs in that Pauli basis. If all four outcomes equal v, prepare|v>; otherwise retain the untouched fifth input. Forget the classical direction and measurement records. This is a randomized trace-preserving stabilizer protocol; classical randomization does not use magic. A uniform three-way choice can be performed classically (or sampling unbiased bits with rejection), and entails no unknown-state-dependent operation.

For fixed pure target Bloch vector r, let S4=r_x^4+r_y^4+r_z^4. Since sum r_a^2=1, Cauchy-Schwarz gives1/3<=S4<=1. The gain over f_t, averaging only the protocol's own classical and measurement randomness, is exactly

G_t(r)=(1/6) sum_v ((1+t v.r)/2)^4 *((v.r-t)/2)
       =[t-6t^3+(4t^3-t^5)S4]/96.

Derivation: pair each axis x with-x and expand
(1+t x)^4(x-t)+(1-t x)^4(-x-t)
=2[-t+(4t-6t^3)x^2+(4t^3-t^5)x^4].
Summing the three axes and dividing by192 gives the displayed gain.

At t=1/3, delta=2/3, the coefficient4t^3-t^5=35/243 is positive. Therefore, uniformly over EVERY pure qubit target,

G_(1/3)(r)>=[1/9+(35/243)/3]/96=29/17496>0,
F_det(r)>=2/3+29/17496=11693/17496>2/3.

The Haar average uses E S4=3/5, and is

F_Haar=2/3+1/486=325/486>2/3.

The output need not be a pure state; fidelity to the pure target is linear in the output, exactly as in the source definition. Successful probability is identically1. We neither condition on nor reveal a favorable target. This addresses even the stronger pointwise interpretation of universal purification, at this stated noise.

For all t with0<t<sqrt(2sqrt(13)-7), the same pointwise guarantee is positive, from3-14t^2-t^4>0. Only the explicit rational t=1/3 point is needed for a counterexample. Discarding additional copies extends to n>=5. Independent acceptance at any fixed0<s<=1 preserves these conditional fidelities with state-independent success. None of this contradicts the two-copy theorem or the project's three-copy no-go proof, and no globally optimal purification fidelity is asserted. Four-copy deterministic purification remains a separate question.

The primary2504.10516v2 Eqs.(2)-(3), AppendixB(S25), and the literal QIQC Eq.(2) permit these maps. The later2607.08626v1 proves two-copy resource tradeoffs only. Source Table1's four-copy numerical statement conflicts with the earlier four-copy example; this five-copy theorem alone need not contradict that finite table. Current public code at QuAIR/NonMagic-Purification-codes/main/Non_magic_purification/universal_cspo_purification_depo_noise.m is configured n=2 and does not establish the four-copy claim. No diagnosis of the authors' unpublished n4 calculation is made. Novelty and external expert confirmation are unverified.

