By Gallier J.
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Example text
Later on, it may be possible to use ∀-introduction without running into trouble with free occurrences of u in the premises. But we still have to be very careful when we use ∀-introduction or ∃-elimination. Before concluding this section, let us give a few more examples of proofs using the rules for the quantifiers. First, let us prove that ∀tP ≡ ∀uP [u/t], where u is any variable not free in ∀tP and such that u is not captured during the substitution. This rule allows us to rename bound variables (under very mild conditions).
This is basically a version of Russell’s paradox. , there is no set to which every other set belongs. Proof . Let A be any set. We construct a set, B, that does not belong to A. If the set of all sets existed, then we could produce a set that does not belong to it, a contradiction. Let B = {a ∈ A | a ∈ / a}. We claim that B ∈ / A. We proceed by contradiction, so assume B ∈ A. However, by the definition of B, we have B∈B iff B ∈ A and B ∈ / B. Since B ∈ A, the above is equivalent to B∈B iff B ∈ / B, which is a contradiction.
Therefore, the set of all natural numbers, N, does exist. The set N is also denoted ω. 4 The set N is inductive and it is a subset of every inductive set. Proof . Recall that ∅ belongs to every inductive set; so, ∅ is a natural number (0). As N is the set of natural numbers, ∅ (= 0) belongs to N. , n + 1 ∈ N. Since N is the set of natural numbers and since every natural number belongs to every inductive set, we conclude that N is a subset of every inductive set. It would be tempting to view N as the intersection of the family of inductive sets, but unfortunately this family is not a set; it is too “big” to be a set.
Discrete Mathematics for Computer Science by Gallier J.
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