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Proof step 1 from proof ext-rec (conclusion)

⊦ st1 = st0[a, d] ↔ a ∈ st1 ∧ st1[a] = d ∧ (∀ b. a ≠ b → (b ∈ st0 ↔ b ∈ st1) ∧ (b ∈ st0 → st0[b] = st1[b]))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
a ∈ st1 ∧ st1[a] = d ∧ (∀ b. a ≠ b → (b ∈ st0 ↔ b ∈ st1) ∧ (b ∈ st0 → st0[b] = st1[b])) ⊦ st1 = st0[a, d]

Rule argument

pos: right 1

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d;%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%At-other: ⊦ a ≠ b → st[b, d][a] = st[a];%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;