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Proof step 2 from proof indef (previous step)

⊦ a ∈ st ↔ (∀ a0. (a0 ∈ st ↔ a0 ∈ st[a, st[a]]) ∧ st[a0] = st[a, st[a]][a0])

User applied rule case distinction

And obtained the following new goals

Proof step 3:
a ∈ st ⊦ ∀ a0. (a0 ∈ st ↔ a0 ∈ st[a, st[a]]) ∧ st[a0] = st[a, st[a]][a0]
Proof step 5:
∀ a0. (a0 ∈ st ↔ a0 ∈ st[a, st[a]]) ∧ st[a0] = st[a, st[a]][a0] ⊦ a ∈ st

Rule argument

pos: right 1

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