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Proof step 5 from proof in-diff (previous step)

⊦ @(m, m0, m1).lnum ∈ st \ ns ∧ @(@(m, m0, m1).pos, @(m, m0, m1).size) ∈ (st \ ns)[@(m, m0, m1).lnum].ladrlist ↔ @(m, m0, m1) ∈ st ∧ ¬ m ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ m ∈ st \ ns ∧ @(m0, m1) ∈ (st \ ns)[m].ladrlist ↔ @(m, m0, m1) ∈ st ∧ ¬ m ∈ ns

Rule argument

Additional information

Used simplifier rules: ⊦ @(m, m0, m1).size = m1; (s)% ⊦ @(m, m0, m1).pos = m0; (s)% ⊦ @(m, m0, m1).lnum = m; (s)