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Proof step 7 from proof restrict-further (previous step)

adr ∈ as0, adr ∈ as1, adr ∈ iast0, as1 ⊆ as0, ∀ adr. adr ∈ iast0 | as0 → adr ∈ iast1 ∧ (iast0 | as0)[adr] = iast1[adr] ⊦ # as0 = 0, # as1 = 0, iast0[adr] = iast1[adr], as0 = ∅, as1 = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 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