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Proof step 5 from proof notin-del-notin (previous step)

⊦ a ∈ st, (a0 ∈ st -- a → a0 ∈ st) ∧ (st -- a)[a0] = st[a0]

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: Del-notin: ⊦ ¬ a ∈ st -- a;%lookupisempty: ⊦ (st -- a)[a] = (∅)[a];%notmemlookup: ⊦ ¬ a ∈ st → st[a] = (∅)[a];%Del-at: ⊦ a ≠ b → (st -- b)[a] = st[a];%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; Used formulas: a0 = a