Automated Reasoning in Higher Order Logic
Let A and B be objects. Assume A=B. Then we have A⊆B.
The following background is necessary to understand this theorem.
(.) Let x and y be objects. Then x=y is a proposition.
(.) Let A and B be objects. Then "A⊆B" is a proposition.
The following background is necessary for the proof.
(.) Let A and x be objects. Then x∈A 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).
(.) Let A and B be objects. Assume for all a ∈ A a∈B. Then we know A⊆B.
Clearly, A⊆A. Using this and A=B, we conclude A⊆B.
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