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Proof step 7 from proof nodes-alloc (previous step)

⊦ n ∈ st, (adr.lnum ∈ st[n, mkblock([], [], m)] ∧ @(adr.pos, adr.size) ∈ st[n, mkblock([], [], m)][adr.lnum].ladrlist → adr ∈ st) ∧ ( adr.lnum ∈ st[n, mkblock([], [], m)] ∧ @(adr.pos, adr.size) ∈ st[n, mkblock([], [], m)][adr.lnum].ladrlist → st[n, mkblock([], [], m)][adr] = nodes(st)[adr])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
adr = @(m0, m1, m2) ⊦ n ∈ st, (m0 ∈ st[n, mkblock([], [], m)] ∧ @(m1, m2) ∈ st[n, mkblock([], [], m)][m0].ladrlist → adr ∈ st) ∧ (m0 ∈ st[n, mkblock([], [], m)] ∧ @(m1, m2) ∈ st[n, mkblock([], [], m)][m0].ladrlist → st[n, mkblock([], [], m)][adr] = nodes(st)[adr])

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, m0, m1, m2]

Additional information

Used simplifier rules: ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1); (elim) Used terms: adr, m0, m1, m2