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Proof step 5 from proof in-remove-refsize0-refsize-cons (previous step)

inv(st), adr ∈ st, adr ∈ as, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, adr.lnum ∈ remove-refsize0(st) ∧ @(adr.pos, adr.size) ∈ remove-refsize0(st)[adr.lnum].ladrlist

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
adr = @(m, m0, m1), inv(st), adr ∈ as, adr ∈ st, ∀ n. n ∈ st → st[n].ref-size = (as | n).size ⊦ # as = 0, as = ∅, m ∈ remove-refsize0(st) ∧ @(m0, m1) ∈ remove-refsize0(st)[m].ladrlist

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: ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1); (elim) Used terms: adr, m, m0, m1