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Can a numeric matrix recover a symbolic rule?

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On printed page seven, the paper measures the implication C M between bra X and ket Y,

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leaving both inputs as variables. We will use the same AND-and-exclusive-or arithmetic

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as a numeric measurement, then simplify the intermediate
expressions to recover the whole implication rule.

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Bra X is the row X, not X. Ket Y is the column Y above not Y.

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The implication C M stays numeric: one, zero above one, one.

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Only the state vectors contain variables at this stage.

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For the first intermediate entry, pair X and not X with the first column, one and one.

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This gives X exclusive-or not X.

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Exactly one of a Boolean value and its negation is true, so that entry

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Exactly one of a Boolean value and its negation is true, so that entry simplifies to one.

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The second column is zero above one. Pairing it with the same bra gives X and zero,

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exclusive-or not X and one. The

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exclusive-or not X and one. The first term vanishes; the

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exclusive-or not X and one. The first term vanishes; the second becomes not X.

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We now know both entries of the intermediate row.

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The paper writes the row as X exclusive-or not X, followed by not X.

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We can also write it as one, not X. These two rows are equal.

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Keep ket Y on the right, because the final contraction still has to be done.

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Multiply the first row entry by Y and the second by not Y, then

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Multiply the first row entry by Y and the second by not Y, then exclusive-or the products.

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The result is Y exclusive-or not X and not Y.

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The parentheses matter: the second term requires both not X and not Y.

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These two terms cannot both be true. If Y is true, the term containing not Y is false.

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If that second term is true, Y is false.

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For these disjoint terms, exclusive-or and OR give the same result.

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That replacement is not valid for arbitrary overlapping terms.

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So we have Y or the conjunction of not X and not Y.

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Distribute OR over that conjunction. One factor becomes Y or not Y, which is always true.

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The remaining factor is not X or Y: precisely X implies Y.

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Here is the derivation together, on separate lines so each step stays readable.

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The numeric C M and the two symbolic vectors produce the intermediate row, then the expression,

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then implication. No valuation of X or Y was needed to establish this identity.

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Use the supplied input X equal to one and Y equal to zero. First evaluate not X or Y.

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Then use the true-first axes to locate the same input in the implication matrix.

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Check that the two methods agree before revealing the result.

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Not X is zero and Y is zero, so their OR is zero.

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In the true-first matrix, X selects the first row and Y selects the second column.

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The entry there is zero. This is the one case in which implication fails.

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In this lesson the C M was numeric, while its state vectors contained expressions.

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Next we move expressions into the matrix cells themselves. That is the logical-matrix layer,

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and we will keep it visually distinct from the numeric operator matrices you now know how to use.