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Proof step 3 from proof at-write-same (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
adr = @(m0, m1, m2), @(m1, m2) ∈ st[m0].ladrlist, m0 ∈ st, n ∈ st ⊦ st[n, mkblock(st[n].list, st[n].ladrlist, m)][adr] = st[m0].list[pos(@(m1, m2), st[m0].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, m2, m3, m4]

Additional information

Used simplifier rules: ⊦ (m = adr.lnum ∧ m0 = adr.pos) ∧ m1 = adr.size ↔ adr = @(m, m0, m1); (elim) Used terms: adr, m2, m3, m4