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Proof step 16 from proof size-eq-0 (previous step)

adr = @(m2, m3, m4), adr ∈ as0, ∀ adr0. adr0 = @(m, m0, m1) ∨ adr0 ∈ as0 → adr0.size = 0, ∀ as1. # as1 < #(as0 ++ @(m, m0, m1)) → true → (0 = as1.size ↔ (∀ adr. adr ∈ as1 → adr.size = 0)) ⊦ # as0 = 0, 0 = as0.size + m1, 0 = as0.size, as0 = ∅, m4 = 0, @(m, m0, m1) ∈ as0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
@(m2, m3, m4) ∈ as0, ∀ adr0. adr0 = @(m, m0, m1) ∨ adr0 ∈ as0 → adr0.size = 0, ∀ as1. # as1 < #(as0 ++ @(m, m0, m1)) → true → (0 = as1.size ↔ (∀ adr. adr ∈ as1 → adr.size = 0)) ⊦ # as0 = 0, 0 = as0.size, 0 = as0.size + m1, as0 = ∅, m4 = 0, @(m, m0, m1) ∈ as0

Proof step 16 from proof size-eq-0 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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