Derive the paper’s implication example

Can a numeric matrix recover a symbolic rule? · 0:00

Can a numeric matrix recover a symbolic rule? On printed page seven, the paper measures the implication C M between bra X and ket Y, leaving both inputs as variables. We will use the same AND-and-exclusive-or arithmetic as a numeric measurement, then simplify the intermediate expressions to recover the whole implication rule.

Define the two symbolic vectors · 0:25

Bra X is the row X, not X. Ket Y is the column Y above not Y. The implication C M stays numeric: one, zero above one, one. Only the state vectors contain variables at this stage.

Multiply into the first column · 0:40

For the first intermediate entry, pair X and not X with the first column, one and one. This gives X exclusive-or not X. Exactly one of a Boolean value and its negation is true, so that entry simplifies to one.

Multiply into the second column · 0:55

The second column is zero above one. Pairing it with the same bra gives X and zero, exclusive-or not X and one. The first term vanishes; the second becomes not X. We now know both entries of the intermediate row.

The intermediate row in the paper · 1:11

The paper writes the row as X exclusive-or not X, followed by not X. We can also write it as one, not X. These two rows are equal. Keep ket Y on the right, because the final contraction still has to be done.

Contract with the ket · 1:28

Multiply the first row entry by Y and the second by not Y, then exclusive-or the products. The result is Y exclusive-or not X and not Y. The parentheses matter: the second term requires both not X and not Y.

Why XOR can become OR here · 1:45

These two terms cannot both be true. If Y is true, the term containing not Y is false. If that second term is true, Y is false. For these disjoint terms, exclusive-or and OR give the same result. That replacement is not valid for arbitrary overlapping terms.

Simplify to the familiar rule · 2:05

So we have Y or the conjunction of not X and not Y. Distribute OR over that conjunction. One factor becomes Y or not Y, which is always true. The remaining factor is not X or Y: precisely X implies Y.

Read the complete chain · 2:24

Here is the derivation together, on separate lines so each step stays readable. The numeric C M and the two symbolic vectors produce the intermediate row, then the expression, then implication. No valuation of X or Y was needed to establish this identity.

Guided consistency check: implication at one input · 2:42

Use the supplied input X equal to one and Y equal to zero. First evaluate not X or Y. Then use the true-first axes to locate the same input in the implication matrix. Check that the two methods agree before revealing the result.

Prediction pause: 10 seconds.

Answer: the one rejected input pair · 3:09

Not X is zero and Y is zero, so their OR is zero. In the true-first matrix, X selects the first row and Y selects the second column. The entry there is zero. This is the one case in which implication fails.

The next layer: expressions inside the matrix · 3:24

In this lesson the C M was numeric, while its state vectors contained expressions. Next we move expressions into the matrix cells themselves. That is the logical-matrix layer, and we will keep it visually distinct from the numeric operator matrices you now know how to use.