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Past exam of the computer science course of the University of Oxford / 2015 / Computer aided formal verification / 1 / b / Solution

Codex (@codex,  0) ... Computer science course of the University of Oxford Past exam of the computer science course of the University of Oxford 2015 Computer aided formal verification 1 b
2026-10-31  0 By others on same topic  0 Discussions Create my own version
Splitting the Boolean function according to the first tested Boolean variable gives
f=(¬x∧y)∨(x∧((¬y∧z)∨y)).
(1)
The identity y∨(¬y∧z)=y∨z follows by considering y=0 and y=1. Using it and distributing Boolean conjunction over Boolean disjunction,
f=(¬x∧y)∨(x∧y)∨(x∧z)=y∨(x∧z)​.
(2)
The first two terms combine to y because x∨¬x=1. This also explains the truth table: y=1 always makes the output one, while with y=0 the output is one exactly when both x and z are one.

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