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Proof step 1 from proof btree-sub-clean-0 (conclusion)

⊦ btree-sub-clean(r, 0, zh) ↔ r ∈ zh

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ btree-sub-clean(r, n, zh) ↔ r ∈ zh ∧ ( ¬ zh[r].leaf → (∀ n0. n0 < zh[r].usedsize ∧ n0 < n ∧ zh[r].zbranches[n0].branch? ∧ zh[r].zbranches[n0].child ∈ zh → ¬ zh[zh[r].zbranches[n0].child].dirty))

Rule argument

spec: <toplevel> name: btree-sub-clean seq: ⊦ btree-sub-clean(r, n, zh) ↔ r ∈ zh ∧ ( ¬ zh[r].leaf → (∀ n0. n0 < zh[r].usedsize ∧ n0 < n ∧ zh[r].zbranches[n0].branch? ∧ zh[r].zbranches[n0].child ∈ zh → ¬ zh[zh[r].zbranches[n0].child].dirty)) subst: [zh, n, r] → [zh, 0, r] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ m < 0; (s) Used terms: zh, 0, r