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Typo - invariant should be AnswerIs(n_0!)

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WZY 2016-03-09 11:02:24 -05:00
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frap_book.tex
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@ -873,7 +873,7 @@ Another important property of invariants formalizes the connection with weakenin
\end{theorem} \end{theorem}
Note that the larger $I'$ above may not be suitable to use in an inductive proof by Theorem \ref{invariant_induction}! Note that the larger $I'$ above may not be suitable to use in an inductive proof by Theorem \ref{invariant_induction}!
For instance, for factorial, we might define $I' = \mathcal \{\mathsf{AnswerIs}(n_0)\} \cup \{\mathsf{WithAccumulator}(n, a) \mid n, a \in \mathbb N\}$, clearly a superset of $I$. For instance, for factorial, we might define $I' = \mathcal \{\mathsf{AnswerIs}(n_0!)\} \cup \{\mathsf{WithAccumulator}(n, a) \mid n, a \in \mathbb N\}$, clearly a superset of $I$.
However, by forgetting everything that we know about intermediate $\mathsf{WithAccumulator}$ states, we will get stuck on the inductive step of the proof. However, by forgetting everything that we know about intermediate $\mathsf{WithAccumulator}$ states, we will get stuck on the inductive step of the proof.
Thus, what we call invariants here needn't also be \emph{inductive invariants}\index{inductive invariants}, and there may be slight terminology mismatches with other sources. Thus, what we call invariants here needn't also be \emph{inductive invariants}\index{inductive invariants}, and there may be slight terminology mismatches with other sources.