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Proof step 7 from proof ext-rec (previous step)

a0 ≠ a, a ∈ st1, ∀ b. a ≠ b → (b ∈ st0 ↔ b ∈ st1) ∧ (b ∈ st0 → st0[b] = st1[b]) ⊦ (a0 ∈ st1 ↔ a0 ∈ st0[a, st1[a]]) ∧ st1[a0] = st0[a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ st1, ∀ b. a ≠ b → (b ∈ st0 ↔ b ∈ st1) ∧ (b ∈ st0 → st0[b] = st1[b]) ⊦ a0 = a, (a0 ∈ st1 ↔ a0 ∈ st0) ∧ st1[a0] = st0[a0]

Rule argument

Additional information

Used simplifier rules: in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);