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Proof step 4 from proof Data-notin-other (previous step)

a0 ∈ st[a, d1] ⊦ st[a, d1][a0] = d1, a0 = a, ∃ a1. a1 ∈ st ∧ st[a1] = st[a, d1][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ∈ st ⊦ st[a0] = d1, a0 = a, ∃ a1. a1 ∈ st ∧ st[a1] = st[a0]

Rule argument

Additional information

Used simplifier rules: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st); (s)% ⊦ a ≠ b → st[b, d][a] = st[a]; (s)