# Dimensional Chess: Initial-Position Mate Frontier R1

**Version:** 1.0.0  
**Date:** 15 September 2026  
**Status:** Independently verified seven-ply bounded result; not a chess solution

## Result

The C6 dependency search was run on the actual orthodox initial position:

```text
rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1
```

The exact bounded AND/OR predicate returns false for both possible attackers at
depth seven:

\[
B_W(s_0,7)=0,
\qquad
B_B(s_0,7)=0.
\tag{R1.1}
\]

Therefore neither White nor Black can force checkmate from the initial position
within seven plies. This does not determine whether perfect chess is a White
win, draw, or Black win.

## 1. C6 dependency runs

| Attacker | Result | Nodes | Generated edges | Horizon nodes | Exchange graphs | Time |
|---|---:|---:|---:|---:|---:|---:|
| White | no forced mate ≤7 | 128,724 | 251,799 | 78,842 | 5,370 | 105.820 s |
| Black | no forced mate ≤7 | 14,846 | 143,274 | 4,046 | 3,127 | 53.319 s |

The asymmetry is expected. White is the side to move. To disprove a White
forced mate, every candidate White first move must be defeated. To disprove a
Black forced mate, White needs only one root continuation that defeats the
universal Black-mate claim.

For White as attacker, the dependency search grew as follows:

| Horizon | Nodes | Time |
|---:|---:|---:|
| 1 ply | 21 | 0.011 s |
| 3 plies | 486 | 0.381 s |
| 5 plies | 7,536 | 6.873 s |
| 7 plies | 128,724 | 105.820 s |

The last two-ply extension multiplies nodes by approximately 17.08. The current
method has not changed the exponential character of the frontier.

## 2. Independent lexical verification

A separate verifier imports no C6 code and uses lexical UCI move order. It
independently recomputes the same bounded predicate.

| Attacker | Independent result | Lexical nodes | Lexical edges |
|---|---:|---:|---:|
| White | no forced mate ≤7 | 106,322 | 230,521 |
| Black | no forced mate ≤7 | 15,082 | 131,361 |

Both results agree with C6. The verifier reports `PASS`.

The comparison also falsifies a stronger ordering claim:

- for the White-attacker proof, C6 uses 21.07% more nodes than lexical order;
- for the Black-attacker proof, C6 uses 1.56% fewer nodes;
- combined, C6 uses 143,570 nodes versus 121,404, or 18.26% more.

C6's aggregate advantage on the eight-position tactical corpus therefore does
not transfer to the initial-position negative-mate proof.

## 3. What was proved

The finite proposition in Equation R1.1 is exact relative to the verified move
engine. Each search uses every legal orthodox successor required by the AND/OR
predicate. The exchange dependency value changes traversal order only; it
cannot change the Boolean result. An independent implementation with a
different order reaches the same answer for both attackers.

At this horizon, the initial halfmove clock and repetition history cannot reach
the automatic-draw thresholds. Thus omitting a draw claim cannot turn a forced
mate into the reported negative result.

## 4. What was not proved

The run does not establish:

- a game-theoretic value for the initial position;
- absence of a forced mate beyond seven plies;
- a lower or upper WDL interval narrower than `[-1,+1]`;
- a sound move deletion at the root;
- polynomial scaling; or
- a solution of chess.

The result is also not evidence that lexical order is generally superior.
Lexical order has no chess semantics; it happened to encounter counterexample
branches earlier in this negative proof.

## 5. Consequence: ordering is not the wallbreaker

Positive and negative AND/OR proofs demand different witnesses:

- at an attacker node, proving mate needs one successful move while disproving
  mate needs every move to fail;
- at a defender node, proving mate needs every reply covered while disproving
  mate needs one escape.

A fixed notion of “promising move” therefore cannot be optimal for both proof
directions. C5 and C6 improve computation allocation in some tactical searches,
but they do not create information about unexpanded subtrees.

The next stage must attach sound intervals to forcing-dependency graph nodes.
The target is a dual certificate:

\[
I(s)=[D^-(s),D^+(s)]
\tag{R1.2}
\]

with proof-carrying rules for checks, mandatory evasions, legal exchanges,
promotions, quiet threats, repetition, and draw claims. Only a bound such as

\[
D^-(b)\ge D^+(a)
\tag{R1.3}
\]

can delete action $a$ in favor of $b$ at a maximizing node. Without those
bounds, deeper C6 runs only repeat the exponential expansion at a larger cost.

The R1 run therefore advances the chess program by locating the wall precisely:
the missing object is not another scalar heat score. It is a compact,
transferable, independently checkable bound on all completions represented by
an abstract dependency state.