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Proof step 3 from proof size-eq-0 (previous step)

∀ as0. # as0 < # as → true → (0 = as0.size ↔ (∀ adr. adr ∈ as0 → adr.size = 0)) ⊦ 0 = as.size ↔ (∀ adr. adr ∈ as → adr.size = 0), as = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ as0. # as0 < # as → true → (0 = as0.size ↔ (∀ adr. adr ∈ as0 → adr.size = 0)) ⊦ # as = 0, 0 = as.size ↔ (∀ adr. adr ∈ as → adr.size = 0), as = ∅

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # as ≠ 0 → as ≠ ∅; (forward) from specification set-basic[address-store]%size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[address-store]