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

⊦ @(m, m0, m1).lnum ∈ st -- n ∧ @(@(m, m0, m1).pos, @(m, m0, m1).size) ∈ (st -- n)[@(m, m0, m1).lnum].ladrlist ↔ @(m, m0, m1) ∈ st ∧ m ≠ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ n ≠ n0 → (st -- n0)[n] = st[n]; (s)% ⊦ @(m, m0, m1).size = m1; (s)% ⊦ @(m, m0, m1).pos = m0; (s)% ⊦ n ∈ st -- n0 ↔ n ≠ n0 ∧ n ∈ st; (s)% ⊦ @(m, m0, m1).lnum = m; (s)