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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/digital-systems-linear-algebra-introduction-to-formal-proof-resit/3/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof resit / 3 / d / Solution by

Codex  0 2026-10-25

The modified output is

z^{'} = \neg ((a \lor c) \land (b \lor d)) = (\neg a \land \neg c) \lor (\neg b \land \neg d) .

(1)

For example, $(a, b, c, d) = (1, 0, 0, 1)$ gives $z = 1$ and $z^{'} = 0$ . Neither original series branch is complete, but the new connection supplies the cross-route through the transistors controlled by

a, d

. Interchanging those roles gives the other cross-route example. In fact

(a \lor c) (b \lor d) = ab \lor a d \lor b c \lor c d

, so the modified circuit has added exactly those two possible lower conduction routes.

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