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Proof step 5 from proof btree-sub-clean-next (previous step)

btree-sub-clean(r, n, zh) ⊦ r ∈ zh ∧ ( ¬ zh[r].leaf → (∀ n0. n0 < zh[r].usedsize ∧ n0 < n + 1 ∧ zh[r].zbranches[n0].branch? ∧ zh[r].zbranches[n0].child ∈ zh → ¬ zh[zh[r].zbranches[n0].child].dirty)) ↔ ¬ zh[r].leaf ∧ n < zh[r].usedsize ∧ zh[r].zbranches[n].branch? ∧ zh[r].zbranches[n].child ∈ zh → ¬ zh[zh[r].zbranches[n].child].dirty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
btree-sub-clean(r, n, zh) ⊦ r ∈ zh ∧ ( ¬ zh[r].leaf → (∀ n0. n0 < zh[r].usedsize ∧ ¬ n < n0 ∧ zh[r].zbranches[n0].branch? ∧ zh[r].zbranches[n0].child ∈ zh → ¬ zh[zh[r].zbranches[n0].child].dirty)) ↔ ¬ zh[r].leaf ∧ n < zh[r].usedsize ∧ zh[r].zbranches[n].branch? ∧ zh[r].zbranches[n].child ∈ zh → ¬ zh[zh[r].zbranches[n].child].dirty

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)