Theorem

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 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 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 x and y be objects. Then x=y is a proposition.

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

(.) Let x and y be objects. Let φ(x) be a proposition depending on an object x. Assume x=y and φ(x). Then we know φ(y).

(.) ∅ is an object.

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then {a∈A|φ(a)} is an object.

(.) 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∈A|φ(a)}.

(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then ∃a∈A⋅φ(a) is the proposition given by ¬{a∈A|φ(a)}=∅.

(.) "false" is a proposition.

(.) Let x be an object. Assume x∈∅. Let φ be a proposition. Then we know φ.

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

Proof

It is enough to show ¬{a∈A|φ(a)}=∅. We will show {a∈A|φ(a)}=∅ implies false. Assume {a∈A|φ(a)}=∅. It is enough to show a∈∅. Since φ(a), we know a∈{a∈A|φ(a)}. Using this and {a∈A|φ(a)}=∅, we are done.

Click here to modify variable names (JavaScript Must Be Enabled).

Hide Background.

Test yourself on this item.