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Proof step 2 from proof in-diff-explicit (previous step)

⊦ @(n0, n1, n2) ∈ st \ ns ↔ @(n0, n1, n2) ∈ st, n0 ∈ ns

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in-diff: ⊦ adr ∈ st \ ns ↔ adr ∈ st ∧ ¬ adr.lnum ∈ ns;% ⊦ @(m, m0, m1).lnum = m; (s)