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Proof step 5 from proof abs-fold-del (previous step)

# entries = m + 1, tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(m, md, m0, m1, entries -- str) * abs(sdt -- str, entries -- str)))(dirs), p ∈ tree, str ∈ entries ⊦ entries = ∅, abs(tree -- (p + str '), ROOT_INO)(dirs)

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ str ∈ entries ∧ p ∈ tree ∧ tree[p] = dnode(md, sdt) ∧ ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(# entries - 1, md, m0, m1, entries -- str) * abs(sdt -- str, entries -- str) * D))(dirs) → ((abs(tree -- (p + str '), ROOT_INO) * D))(dirs)

Rule argument

spec: <toplevel> name: abs-fold-del-p seq: ⊦ str ∈ entries ∧ p ∈ tree ∧ tree[p] = dnode(md, sdt) ∧ (( abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir (# entries - 1, md, m0, m1, entries -- str) * abs(sdt -- str, entries -- str) * D) )(dirs) → ((abs(tree -- (p + str '), ROOT_INO) * D))(dirs) subst: [str, entries, p, tree, md, sdt, ino0, m0, m1, D, dirs] → [str, entries, p, tree, md, sdt, ino0, m0, m1, emp, dirs]

Additional information

Used simplifier rules: ⊦ emp * D = D; (s)%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: str, entries, p, tree, md, sdt, ino0, m0, m1, emp, dirs