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Proof step 9 from proof in-diff (previous step)

(st \ ns)[m] = mkblock(x, lax, m2) ⊦ m ∈ st \ ns ∧ @(m0, m1) ∈ lax ↔ (@(m, m0, m1).lnum ∈ st ∧ @(@(m, m0, m1).pos, @(m, m0, m1).size) ∈ st[@(m, m0, m1).lnum].ladrlist) ∧ ¬ m ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(st \ ns)[m] = mkblock(x, lax, m2) ⊦ m ∈ st \ ns ∧ @(m0, m1) ∈ lax ↔ m ∈ st ∧ @(m0, m1) ∈ st[m].ladrlist ∧ ¬ 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)