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Proof step 188 from proof refine-create (
previous step
)
abs(tree0, ino1) = emp ⊦ ¬ str ∈ entries ∧ p ∈ tree ∧ tree[p] = dnode(md, sdt) ∧ ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(# entries + 1, md, m0, m1, entries[str, ino1]) * abs(sdt, entries)))(dirs) → abs(tree[p + str ', tree0], ROOT_INO)(dirs)
Lemma
abs-fold-new