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Proof step 5 from proof restrict-add-at (previous step)

adr ∈ iast0, adr ∈ as, ∀ adr. adr ∈ iast0 | as → adr ∈ iast1 ∧ (iast0 | as)[adr] = iast1[adr] ⊦ # as = 0, iast1[adr] = iast0[adr], as = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: adr

Additional information

Used simplifier rules: ⊦ adr ∈ iast | as ↔ adr ∈ iast ∧ adr ∈ as; (ws)% ⊦ adr ∈ iast ∧ adr ∈ as → (iast | as)[adr] = iast[adr]; (s) Used terms: adr