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Proof step 3 from proof notmemlookup (previous step)

∀ a. st0[a] = (∅)[a] ∨ a ∈ st0 ⊦ st0[a0, d][a] = (∅)[a], a ∈ st0[a0, d]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. st0[a] = (∅)[a] ∨ a ∈ st0 ⊦ a = a0, st0[a] = (∅)[a], a ∈ st0

Rule argument

Additional information

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