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Proof step 10 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), p ∈ tree, str ∈ entries, p ∈ tree[p, dnode(md, sdt -- str)] ⊦ entries = ∅, ((_∃(λ ino1. abs|(tree[p, dnode(md, sdt -- str)], ROOT_INO, p, ino1) * abs(tree[p, dnode(md, sdt -- str)][p], ino1)) * D))(dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# 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 = ∅, ∃ ino. (((abs|(tree[p, dnode(md, sdt -- str)], ROOT_INO, p, ino) * abs(dnode(md, sdt -- str), ino)) * D))(dirs)

Rule argument

Additional information

Used simplifier rules: sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]% ⊦ ((_∃ nD * D))(dirs) ↔ (∃ ino. ((nD(ino) * D))(dirs)); (s)% ⊦ p ∈ tree → tree[p, tree0][p] = tree0; (s)% ⊦ p ∈ tree → p ∈ tree[p, tree0]; (s)