Theorem

Let A be an object. Let φ(a) be a proposition depending on a member a of A. Assume ¬(∃a∈A⋅φ(a)). Then we have ∀a∈A⋅¬φ(a).

Background

The following background is necessary to understand this theorem.

(.) Let A and x be objects. Then x∈A is a proposition.

(.) Let φ be a proposition. Then ¬φ is a proposition.

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then "∀a∈A⋅φ(a)" is a proposition.

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then "∃a∈A⋅φ(a)" is a proposition.

Background for Proof

The following background is necessary for the proof.

(.) Let φ and ψ be propositions. Assume φ and ¬φ. Then we know ψ.

(.) "false" is a proposition.

(.) Let φ be a proposition. Assume φ implies false. Then we know ¬φ.

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Assume for all a ∈ A φ(a). Then we know ∀a∈A⋅φ(a).

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Let a be a member of A. Assume φ(a). Then we know ∃a∈A⋅φ(a).

Proof

Let a be a member of φ. It is enough to show φ(a) implies false. Assume φ(a). Hence ∃a∈A⋅φ(a). Using this and ¬(∃a∈A⋅φ(a)), we conclude false.

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