WEBVTT

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How can we store every answer to an if-then rule?

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Our rule says: if W and X differ, then Y must be false and Z must be true.

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The left condition is exclusive-or; the right is NOT Y AND Z.

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We will combine their matrices to construct all sixteen
answers, then check one cell against the original rule.

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Call the two conditions P and Q for a moment.

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P implies Q fails only when P is one and Q is zero.

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If the requirement is triggered and not met, the whole rule is false. Every other pair is accepted.

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After positive valuation, the first condition has the exclusive-or matrix, which we call A.

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The second has matrix B, with its one at Y false and Z true.

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These are separate two-by-two matrices for the two parts of the rule.

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The first accepted case has both conditions true.

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The tensor A with B marks exactly those four-variable assignments.

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Each one indicates a place where the left condition and the right condition both return one.

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The second case has the left condition false and the right condition true.

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Flip every bit of A, then tensor that complement with B.

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Complement here means entrywise Boolean NOT:
every zero becomes one, and every one becomes zero.

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The third accepted case has both conditions false.
Tensor the complement of A with the complement of B.

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The three accepted cases cannot overlap, because each
assignment gives only one pair of values for P and Q.

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Combine the three tensors entry by entry using exclusive-or. Since they do not overlap,

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each accepted assignment keeps one one. This produces the four-by-four matrix from the paper.

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Its displayed row components are W, Y, and its column components are X, Z.

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For W, X, Y, Z equal to zero, one, one, one, evaluate P and

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Q from their definitions. Then decide the implication
and locate the matching row and column in the matrix.

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Explain why the direct rule calculation and the addressed entry agree.

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W and X differ, so P is one. But Y is true, so not Y and Z is zero.

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A true condition implies a false requirement, which fails. The expected output is zero.

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The row uses W and Y, giving zero, one. The column uses X and Z,

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giving one, one. In our descending order this selects the third physical row and the first column.

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Their intersection is zero, agreeing with the direct calculation.

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The paper writes its row selector with the bra factors in reversed written order, Y then W.

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That notation needs its component expansion.

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Our explicit row labels keep W and Y visible in the physical order used by the displayed matrix.

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You have now constructed the matrix and checked an entry without trusting the picture alone.

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Next we will use the repository layout convention,

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where zero comes first and binary input groups become ordinary numeric addresses.