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Proof step 19 from proof size-eq-0 (previous step)

0 = as0.size, ∀ adr. adr ∈ as0 → adr.size = 0, ∀ as1. # as1 < #(as0 ++ @(m, m0, m1)) → true → (0 = as1.size ↔ (∀ adr. adr ∈ as1 → adr.size = 0)) ⊦ m1 = 0 ↔ (∀ adr0. adr0 = @(m, m0, m1) ∨ adr0 ∈ as0 → adr0.size = 0), @(m, m0, m1) ∈ as0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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