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Proof step 23 from proof size-eq-0 (previous step)

0 = as0.size, @(m1, m2, m3) ∈ as0, ∀ adr. adr ∈ as0 → adr.size = 0, ∀ as1. # as1 < #(as0 ++ @(m, m0, 0)) → true → (0 = as1.size ↔ (∀ adr. adr ∈ as1 → adr.size = 0)) ⊦ # as0 = 0, as0 = ∅, m3 = 0, @(m, m0, 0) ∈ as0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: @(m1, m2, m3)

Additional information

Used simplifier rules: ⊦ @(m, m0, m1).size = m1; (s) Used terms: @(m1, m2, m3)