Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Digital systems, linear algebra, introduction to formal proof / 8 / b / iii / Solution

Introduce a fresh existential witness $v$ with $R(v)$. The universal premise gives $v=0\vee v=1$. In the first case, equality elimination rule gives $R(0)$. In the second, it gives $R(1)$, contradicting $\neg R(1)$; explosion then supplies the common case conclusion $R(0)$. Close both cases and then the witness scope:

HenceFor both uses of equality elimination rule, the formula template is $R(z)$. At the final existential elimination, $v$ is fresh in all three original premises and absent from $R(0)$, so the eigenvariable condition holds. No additional assumption that the constants $0$ and $1$ denote distinct objects is needed.