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Proof step 8 from proof exnotempty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a ∈ st, ∀ a0, d, st0. st ≠ st0[a0, d] ∨ a0 ∈ st0 ⊦

Rule argument

Additional information

Used simplifier rules: In-empty: ⊦ ¬ a ∈ ∅;%notmemlookup: ⊦ ¬ a ∈ st → st[a] = (∅)[a];