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Proposition 16

[27Q]Addition is associative.

Proof ▼

Consider

P (h) ≐ \forall n, m \in ℕ, (n + m) + h = n + (m + h);

Obviously $P (0)$ is true, moreover $P (S h)$ is proven (omitting ” $\forall n, m \in ℕ$ ”) like this

\begin{array}{r} (n + m) + S h = S (n + m) + h = (S n + m) + h \overset{P (n)}{=} \\ = S n + (m + h) = n + S (m + h) = n + (m + S h) \qedhere \end{array}

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Authors: "Mennucci , Andrea C. G." .

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