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Proof step 2 from proof at-diff (previous step)

@(n, m0, m1) ∈ st ⊦ (st \ ns)[@(n, m0, m1)] = st[@(n, m0, m1)], n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ∈ st, @(m0, m1) ∈ st[n].ladrlist ⊦ (st \ ns)[@(n, m0, m1)] = st[n].list[pos(@(m0, m1), st[n].ladrlist)], n ∈ ns

Rule argument

Additional information

Used simplifier rules: at: ⊦ adr ∈ st → st[adr] = st[adr.lnum].list [pos(@(adr.pos, adr.size), st[adr.lnum].ladrlist) ];%in: ⊦ adr ∈ st ↔ adr.lnum ∈ st ∧ @(adr.pos, adr.size) ∈ st[adr.lnum].ladrlist;% ⊦ @(m, m0, m1).size = m1; (s)% ⊦ @(m, m0, m1).pos = m0; (s)% ⊦ @(m, m0, m1).lnum = m; (s)