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Proof step 6 from proof subseteq-dom-nodes-alloc-restrict-empty (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: address-data name: elim seq: ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1) subst: [adr, m, m0, m1] → [adr, m, m0, m1]

Additional information

Used simplifier rules: ⊦ ¬ n ∈ st → ¬ @(n, n0, n1) ∈ st; (s)% ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1); (elim) Used terms: adr, m, m0, m1