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Proof step 6 from proof delete-put-rew (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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