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Proof step 16 from proof abs-fold-del-p (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) * D))(dirs), str ∈ entries, p ∈ tree ⊦ entries = ∅, mkdir(m, md, m0, m1, entries -- str).meta = md ∧ mkdir(m, md, m0, m1, entries -- str).size = # mkdir(m, md, m0, m1, entries -- str).entries ∧ (( ( (ino0 => mkdir(m, md, m0, m1, entries -- str) * abs(sdt -- str, mkdir(m, md, m0, m1, entries -- str).entries)) * abs|(tree[p, dnode(mkdir(m, md, m0, m1, entries -- str).meta, sdt -- str)], ROOT_INO, p, ino0)) * D) )(dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# 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) * D))(dirs), str ∈ entries, p ∈ tree ⊦ entries = ∅, (((ino0 => mkdir(m, md, m0, m1, entries -- str) * abs(sdt -- str, entries -- str) * abs|(tree[p, dnode(md, sdt -- str)], ROOT_INO, p, ino0)) * D) )(dirs)

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% ⊦ n + 0 = n; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]% ⊦ n -1 = n - 1; (s)% ⊦ str ∈ entries → #(entries -- str) = (# entries)-1; (s)% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)% ⊦ mkdir(m, md, m0, m1, entries).size = m; (s)% ⊦ mkdir(m, md, m0, m1, entries).meta = md; (s)