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Proof step 6 from proof k-Del-rec2 (previous step)

⊦ a0 = b, a = b, a ∈ st, (a0 ∈ st[a, d] -- b ↔ a0 ∈ (st -- b)[a, d]) ∧ (st[a, d] -- b)[a0] = (st -- b)[a, d][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ a0 = b, a = b, a ∈ st, (a0 ∈ st[a, d] ↔ a0 ∈ (st -- b)[a, d]) ∧ st[a, d][a0] = (st -- b)[a, d][a0]

Rule argument

Additional information

Used simplifier rules: Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%Del-at: ⊦ a ≠ b → (st -- b)[a] = st[a];