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Proof step 3 from proof in-alloc (previous step)

adr.lnum ∈ st, @(adr.pos, adr.size) ∈ st[adr.lnum].ladrlist ⊦ n ∈ st, adr ∈ st[n, ablock]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
adr = @(m, m0, m1), @(m0, m1) ∈ st[m].ladrlist, m ∈ st ⊦ adr ∈ st[n, ablock], n ∈ st

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