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Proof step 8 from proof exnotempty (previous step)

⊦ ∀ a. (a ∈ st ↔ a ∈ ∅) ∧ st[a] = (∅)[a], ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0, ¬ a ∈ st ∧ st[a] = (∅)[a]

Rule argument

Additional information

Used simplifier rules: In-empty: ⊦ ¬ a ∈ ∅;