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Proof step 19 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 ⊦ 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), p ≠ ε ∧ p ⊑ p ∧ parent(p) ∈ tree ∧ tree[parent(p)].dir?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# 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), p ≠ ε ∧ parent(p) ∈ tree ∧ tree[parent(p)].dir?

Rule argument

Additional information

Used simplifier rules: ⊦ p ⊑ p; (s)