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Proof step 2 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)

User applied rule cut formula

And obtained the following new goals

Proof step 3:
∀ as0. # as0 < # as → true → (0 = as0.size ↔ (∀ adr. adr ∈ as0 → adr.size = 0)) ⊦ 0 = as.size ↔ (∀ adr. adr ∈ as → adr.size = 0), as = ∅

Rule argument

fma: as = ∅

Additional information

Used simplifier rules: size-emp: ⊦ ∅.size = 0;% ⊦ ¬ adr ∈ ∅; (s) Used formulas: as = ∅