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Proof step 5 from proof subseteq-dom-nodes-alloc-restrict-empty (previous step)

as ⊆ dom nodes(st) ⊦ #(as | n) = 0, n ∈ st, ∀ adr. adr ∈ as | n ↔ adr ∈ ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
adr ∈ st, adr ∈ as, as ⊆ dom nodes(st) ⊦ # as = 0, #(as | adr.lnum) = 0, as = ∅, as | adr.lnum = ∅, adr.lnum ∈ st

Rule argument

Additional information

Used simplifier rules: restrict-leb: ⊦ adr ∈ as | n ↔ adr.lnum = n ∧ adr ∈ as;%size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[address-store]%size-zero: ⊦ # as ≠ 0 → as ≠ ∅; (forward) from specification set-basic[address-store]% ⊦ adr ∈ nodes(st) ↔ adr ∈ st; (s)% ⊦ adr ∈ dom ast ↔ adr ∈ ast; (s)%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[address-store]%sub-in: ⊦ adr ∈ as ∧ as ⊆ as0 → adr ∈ as0; (forward) from specification set-basic[address-store]% ⊦ ¬ adr ∈ ∅; (s)