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Proof step 2 from proof subset-put (previous step)

⊦ (∀ a0. a0 ∈ st[a, d] → a0 ∈ st0 ∧ st[a, d][a0] = st0[a0]) ↔ (∀ a0. a0 ∈ st -- a → a0 ∈ st0 ∧ st[a0] = st0[a0]) ∧ a ∈ st0 ∧ st0[a] = d

User applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a0. a0 ∈ st[a, d] → a0 ∈ st0 ∧ st[a, d][a0] = st0[a0] ⊦ (∀ a0. a0 ∈ st -- a → a0 ∈ st0 ∧ st[a0] = st0[a0]) ∧ a ∈ st0 ∧ st0[a] = d
Proof step 7:
(∀ a0. a0 ∈ st -- a → a0 ∈ st0 ∧ st[a0] = st0[a0]) ∧ a ∈ st0 ∧ st0[a] = d ⊦ ∀ a0. a0 ∈ st[a, d] → a0 ∈ st0 ∧ st[a, d][a0] = st0[a0]

Rule argument

pos: right 1

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