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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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Toffoli computes into a clean target and uncomputes it

Statement

Let a,b be the values on two distinct control wires, let t be a distinct clean target wire with value 0, and let y be another distinct target wire with arbitrary value. Toffoli changes t to ab while preserving the controls. Copying that value into y by CNOT and applying Toffoli again, in the written chronological order, sends

∣a,b,y,0⟩⟼∣a,b,y⊕ab,0⟩.

The same operator identity holds on superpositions.

Facts & Assumptions

Given: The Toffoli action, a zero-initialized workspace bit, and a distinct arbitrary target bit. Prove the stated compute-copy-uncompute identity.

[F1]

Toffoli and Fredkin gates: Toffoli maps ∣a,b,t⟩ to ∣a,b,t⊕ab⟩.

[F2]
[F3]

Clean and dirty ancillas: A clean ancilla is initialized in ∣0⟩ and the complete subroutine must return it exactly to ∣0⟩.

[F4]

CNOT and Control-Target Order: In the ordered two-qubit basis, CNOT⁡1→2∣a,t⟩=∣a,t⊕a⟩.

[F5]

Denotational semantics of a finite layered circuit: A chronological list of layers L1,…,Ld, with L1 applied first, denotes ULd⋯UL1.

[F6]

Tensor products of finitely many linear maps: A tensor map extends its factor maps linearly and acts factorwise on each elementary tensor.

[F7]

The computational basis is orthonormal: The one-qubit computational basis is an orthonormal basis for its declared inner product.

[F8]

The product basis is orthonormal and tensor dimensions multiply: The ordered tensor product of finite orthonormal factor bases is an orthonormal basis of the register.

Proof

1.1givenF1F3algebra

Since t=0, Toffoli sets t to 0⊕ab=ab. The four control cases 00,01,10,11 give target values 0,0,0,1, respectively, and the controls are unchanged.

2.1givenF1F2F4F5step 1.1algebra

CNOT from this work bit to the disjoint target changes y to y⊕ab and leaves a,b,t otherwise unchanged. The final Toffoli toggles t by the same ab, returning it to ab⊕ab=0; by [F2] it is the inverse compute gate. Thus each basis input has the stated output and clean work.

3.1step 2.1F5F6F7F8algebra∎

The vectors ∣a,b,y,0⟩, as (a,b,y) ranges over its eight values, form a basis of the input subspace with the work factor fixed to ∣0⟩, by [F7]–[F8]. Every gate is linear, so the basis-vector identity in Step 2.1 extends to every superposition in that subspace. The tensor-map rule [F6] makes each gate act factorwise on the ordered wires, preserving the displayed register order.

Depends on

Used by

Dependency tree · two levels

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Sources