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Proof step 2 from proof notin-notin (previous step)

@(n, n0, n1) ∈ st ⊦ n ∈ st

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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