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Proof step 3 from proof rew (previous step)

⊦ a0 = a, a = b, (a0 ∈ st[b, d0][a, d] ↔ a0 ∈ st[a, d][b, d0]) ∧ st[b, d0][a, d][a0] = st[a, d][b, d0][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ a0 = a, a = b, (a0 ∈ st[b, d0] ↔ a0 ∈ st[a, d][b, d0]) ∧ st[b, d0][a0] = st[a, d][b, d0][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];