Separate answer key

Record your attempt before reading this key.

# Independent construction and transfer companions

Attempt the checkpoint exercises before reading the solutions below.

## After lesson 9 — paper/layout companion

**Question.** A system accepts when A or B is true, provided C is false. Write the Boolean rule. Use rows `[A]` and columns `[B,C]` in false-first order. Construct the full 2×4 CM. Predict the result for `ABC=101`, then locate it. Next use rows `[A,B]` and columns `[C]`; show where the same assignment moves. Finally verify every one of the eight assignments independently from the Boolean rule.

**Answer key.** `F=(A∨B)∧¬C`. With rows `[A]`, columns `[B,C]`, the matrix is `[[0,0,1,0],[1,0,1,0]]`; `101` maps to row 1, column 1 and returns 0. With rows `[A,B]`, columns `[C]`, it is `[[0,0],[1,0],[1,0],[1,0]]`; `101` maps to row 2, column 1 and still returns 0. There are three accepted assignments.

## After lesson 23 — executable transfer companion

**Question.** Starting from the same specification, create a fresh expression with the public API, choose a different valid row/column split, materialize it, and compare every entry with a scalar oracle. Then change OR to XOR and repeat the complete check. Explain which cells and count change.

**Answer key.** The XOR variant is `G=(A⇕B)∧¬C`. Under rows `[A]`, columns `[B,C]`, its matrix is `[[0,0,1,0],[1,0,0,0]]` and its accepted-count is 2. Check both rules over all eight assignments with an independent scalar oracle.

## Use in a learner session

Do not show either answer key or the matrix while collecting the first response. Record the rule, chosen layout, address, explanation, time and any hint/replay. A later 24–72 hour XOR transfer attempt can assess retention; no study result is claimed by this package.