Proposition 5.5Matching the opponent's nimber difference restores nim-sum zero

Checked theorem Lean checks a formal statement corresponding to this claim.

1 Paper text

Proposition 5.5 (Local restoration principle) Suppose the opponent moves from a position $Q$ with $\operatorname{nim}(Q)=0$ to a position $Q'$ (so it is now our turn). Let $\Delta(Q,Q')=\operatorname{nim}(Q)\oplus \operatorname{nim}(Q')$ denote the nimber difference caused by the opponent’s move. If we respond with a move $Q' \to R$ such that $\Delta(Q',R) \;=\; \Delta(Q,Q'),$ then $\operatorname{nim}(R)=0$ .

Generated from paper_jmlr.tex, lines 520–529.

2 Explain this

The idea. A NIM position with nim-sum zero is "balanced": whoever must move from it loses against correct play. Any move changes one heap, say from $x_i$ to $x_i'$ , and this unbalances the position by exactly $\delta = x_i \oplus x_i'$ . If the reply changes some heap by the same XOR difference $\delta$ , the two changes cancel and the balance returns.

A small example. In Figure 2, $Q=(3,5,6)$ has nim-sum $3\oplus5\oplus6=0$ . The opponent plays $5\to4$ , so $\delta=5\oplus4=1$ . The reply $3\to2$ also has $3\oplus2=1$ , and $R=(2,4,6)$ has nim-sum $2\oplus4\oplus6=0$ again.

Why it matters. The reply never needs the nim-sum of the whole board. It needs only the two most recent frames: $\delta$ is read off the one heap that changed, and a cancelling heap is found by a local bit test. That is why a shallow two-frame network can play this response although no shallow single-frame network can evaluate the nim-sum (Theorem 5.2).

Caveats.

The promise matters. If $\operatorname{nim}(Q)\neq0$ , matching the difference only reproduces the old value, $\operatorname{nim}(R)=\operatorname{nim}(Q)$ (see the initialisation boundary).
Restoration is conditional: it keeps a won position won once a zero-nim-sum position has been established. It does not find the first winning move from an arbitrary position.
The previous frame must be the board the opponent moved from, that is, the player's own post-move board.

Two-frame restoration, step by step

Exact construction trace Every value is computed from the construction’s definitions for the example shown.

This is the checked two-frame selector run on one example, gate group by gate group. It starts from the two frames of Figure 2 and ends with the binary output fields shown in Figure 7. Every value below is computed from the selector's definition for the boards shown. Choose another example to see several candidates at once, the fallback, or a terminal board. The two examples marked "outside the promise" do not start from a zero-nim-sum board after a legal move, so the guarantee does not apply to them.

Inside the promise. The earlier board has nim-sum 0 and the opponent decreased one heap, so the restoration theorem guarantees that the output restores nim-sum 0.

Earlier board Q (3, 5, 6)nim-sum 0

Current board Q′ (3, 4, 6)nim-sum 1; changed: heap 1

After the reply (2, 4, 6)nim-sum 0; reply in heap 0

Compare the frames bit by bit
Each heap is a 3-bit block. For every heap and every bit position, one two-input XOR compares Q with Q′, all in parallel.
Heap Q Q′ XOR Heap
0 3 011 3 011 000 same
1 5 101 4 100 001 changed
2 6 110 6 110 000 same
Extract the difference δ
Bit k of δ is the OR of column k of the XOR table: 001 = δ = 1.
At most one heap changed, so at most one row is nonzero and the OR equals the XOR of the rows, which is nim(Q) ⊕ nim(Q′).
Propose replacements and test them in parallel
For every heap at once: replace Q′_i by Q′_i ⊕ δ and test whether that is a decrease. The decisive bit is the highest set bit of δ (weight 1): a heap passes exactly when it has that bit set.
Heap Q′ Q′ ⊕ δ Test
0 3 011 2 010 passes: 2 < 3
1 4 100 5 101 fails: 5 > 4
2 6 110 7 111 fails: 7 > 6
Show the bit comparisons
The test x ⊕ δ < x is a fixed-depth circuit: an OR over the possible highest differing bit, requiring equality above it, a 1 in x and a 0 in the replacement at that bit.
- Heap 0: 011 vs 010 — the highest differing bit has weight 1; the heap has 1 there, so the replacement is smaller.
- Heap 1: 100 vs 101 — the highest differing bit has weight 1; the heap has 0 there, so the replacement is larger.
- Heap 2: 110 vs 111 — the highest differing bit has weight 1; the heap has 0 there, so the replacement is larger.
Candidates passing: 1.
Choose between cancelling and the fallback
cancelling move At least one replacement passes, so the flags are the passing tests and the fallback is masked off.
Heap passes nonzero fallback value flag s_i
0 1 1 3 − 1 = 2 1
1 0 1 4 − 1 = 3 0
2 0 1 6 − 1 = 5 0
Encode the first flagged heap as a binary index
For N = 3 heaps the construction uses r = ⌊log₂ N⌋ + 1 = 2 index bits (the encoder itself needs only N ≤ 2^r). The first flag is at heap 0, so the index bits are 00.
Show the priority encoder
f_i = s_i ∧ ¬s_j for all j < i keeps only the first flag; index bit k is the OR of f_i over the heaps i whose binary index has a 1 at bit k. This is a fixed-depth circuit, not a search.
Heap i i in binary s_i first f_i
0 00 1 1
1 01 0 0
2 10 0 0
Index bits (OR of the f_i in each 1-column): 00
Output the move
validity1index (2 bits)00value (3 bits)010
Decoded move: set heap 0 to 2. New board (2, 4, 6), nim-sum 0.

Heap	Q	Q′	XOR	Heap
0	3 011	3 011	000	same
1	5 101	4 100	001	changed
2	6 110	6 110	000	same

Heap	Q′	Q′ ⊕ δ	Test
0	3 011	2 010	passes: 2 < 3
1	4 100	5 101	fails: 5 > 4
2	6 110	7 111	fails: 7 > 6

Heap	passes	nonzero	fallback value	flag s_i
0	1	1	3 − 1 = 2	1
1	0	1	4 − 1 = 3	0
2	0	1	6 − 1 = 5	0

Heap i	i in binary	s_i	first f_i
0	00	1	1
1	01	0	0
2	10	0	0

How this relates to the Lean proof. The selector is Family.restS (B, N): the local difference circuits LocalRestore.dC, the candidate tests of SelectCircuit.varSel and the priority encoder PriorityEncoder.encode, read on the two-frame input LocalRestore.tf. Family.rest_semantics proves that, on the promised domain, these formulas and the compiled network output a legal move that restores nim-sum 0. Outside the promise the circuit still computes the values shown, and no theorem says that the result restores 0.

What this does not show. The selector does not find a first winning move from an arbitrary board. Given a zero-nim-sum predecessor it maintains the invariant (Theorem 5.12(iii)). The boxes of Figure 7 group operations; the stages here group gates, not network layers.

Depicted in Lean: Family.rest_semantics, LocalRestore.restore_equiv, LocalRestore.localDelta_eq, LocalRestore.restC_eval, SelectCircuit.varSel_correct, AttainCircuit.cancel_iff_bit, PriorityEncoder.encode_correct, Family.rest_ac0, Family.rest_net. Explains the paper’s Figure 2, Figure 7.

Computed in your browser by a short re-implementation of the definitions above (tools/visuals/nimvis-core.js). It is not itself checked by Lean. At build time it agreed with Lean's own evaluation of those definitions on all 12,299 reference cases. Lean's evaluation fixes the output wires, the difference δ and the decoded move; the intermediate tables (flags, first-flag bits) are re-derived from the same definitions.

3 How this is formalised

Objects. A board is a list of natural numbers; nimSum folds XOR over it.

The proposition. Writing $q,q',r$ for the nim-sums of $Q,Q',R$ , the paper's proof is the identity $r = q' \oplus \Delta(Q',R) = q'\oplus (q\oplus q') = q$ under the hypothesis $\Delta(Q',R)=\Delta(Q,Q')$ . Lean states it in exactly this form, conditional_restoration: if $q=0$ and $q\oplus q' = q'\oplus r$ then $r=0$ . No legality assumption is needed for this step. one_heap_difference checks the sentence before it: when one heap changes from $a$ to $b$ , the two nim-sums differ by $a\oplus b$ .

What else is checked. The shallow selector mentioned right after the proposition, its network form and the whole-game strategy are in Section 5.3.1, paragraph 2. Without the zero-nim-sum promise, matching deltas does not force zero (Rollout.matching_deltas_do_not_force_zero; see Section 5.4, “The initialisation boundary for restoration”).

Coverage record
Paper anchor	`prop:restore` (manuscript label)
Displayed as	Proposition 5.5
Claim	Matching the opponent's nimber difference restores nim-sum zero
Coverage category	Checked theorem
Statement source	paper_jmlr.tex lines 520–529; SHA-256 of the normalised block source `a8eb6954cec14e23…`
Lean declarations	`NimAudit.conditional_restoration` (NimCore, lines 50–52, file `b07db6c50521…`) `NimAudit.one_heap_difference` (NimCore, lines 37–39, file `b07db6c50521…`)
Assumptions and domain	The predecessor $Q$ has nim-sum zero and $Q\to Q'$ is the opponent's move.
Prose-to-formal notes	The proposition is the algebra in `conditional_restoration`, stated for the three nim-sums $q=\operatorname{nim}(Q)$ , $q'=\operatorname{nim}(Q')$ , $r=\operatorname{nim}(R)$ . Legality of the reply is not needed for this step; the shallow selector in the next sentence is a separate record.
Depends on	Definition 5.3
Audited correspondence rows	`prop:restore`
Explanations	`explain/prop:restore`, `formal/prop:restore`
Build evidence	fresh isolated rebuild, audit report

4 Lean statement and source

Principal declarations

theorem `conditional_restoration`

Module NimCore (lean/NimCore.lean, lines 50–52) · file SHA-256 b07db6c505210d5a… Axioms reported by the fresh build: propext, Quot.sound

Statement as elaborated and printed by Lean 4.26.0

∀ (q q' r : ℕ), q = 0 → q ^^^ q' = q' ^^^ r → r = 0

Exact source text, lines 50–52

50theorem conditional_restoration (q q' r : Nat)51    (zero : q = 0) (matched : q ^^^ q' = q' ^^^ r) : r = 0 := by52  exact (matching_deltas_restore_original q q' r matched).trans zero

Project declarations in the statement: none

Project declarations used by the proof or definition body

matching_deltas_restore_original

theorem `one_heap_difference`

Module NimCore (lean/NimCore.lean, lines 37–39) · file SHA-256 b07db6c505210d5a… Axioms reported by the fresh build: propext, Quot.sound

Statement as elaborated and printed by Lean 4.26.0

∀ (pre post : List ℕ) (a b : ℕ),
  NimAudit.nimSum (pre ++ a :: post) ^^^ NimAudit.nimSum (pre ++ b :: post) = a ^^^ b

Exact source text, lines 37–39

37theorem one_heap_difference (pre post : List Nat) (a b : Nat) :38    nimSum (pre ++ a :: post) ^^^ nimSum (pre ++ b :: post) = a ^^^ b := by39  simp [nimSum_append, nimSum, Nat.xor_comm, xor_left_comm]

Project declarations in the statement: nimSum

Project declarations used by the proof or definition body

nimSum, nimSum_append, xor_left_comm

Definitions these statements use

def `nimSum`

Module NimCore (lean/NimCore.lean, lines 7–9) · file SHA-256 b07db6c505210d5a…

Statement as elaborated and printed by Lean 4.26.0

List ℕ → ℕ

Exact source text, lines 7–9

7def nimSum : List Nat → Nat8  | [] => 09  | h :: t => h ^^^ nimSum t

Project declarations in the statement: none

Project declarations used by the proof or definition body

none

Full modules

Module	File	Lines	Imports
NimCore	`lean/NimCore.lean`	139	`Std`

Build evidence: fresh isolated rebuild of all 38 bundled modules with Lean (version 4.26.0, started 2026-10-10T15:21:56Z, finished 2026-10-10T15:27:00Z: 38/38 modules compiled, 1588 axiom reports, no errors. See Build for the commands and the independent audit of 2026-10-10 (dated evidence).

Proposition 5.5Matching the opponent's nimber difference restores nim-sum zero

1 Paper text

2 Explain this

Two-frame restoration, step by step

Compare the frames bit by bit

Extract the difference δ

Propose replacements and test them in parallel

Choose between cancelling and the fallback

Encode the first flagged heap as a binary index

Output the move

3 How this is formalised

4 Lean statement and source

Principal declarations

theorem conditional_restoration

theorem one_heap_difference

Definitions these statements use

def nimSum

Full modules

theorem `conditional_restoration`

theorem `one_heap_difference`

def `nimSum`