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Proof step 7 from proof absc-path (previous step)

dirs[ino] = mkdir(m, md, m0, m1, entries), ((abs|(tree, ino, str ' + p0, ino0) * D))(dirs), ispath(p0, entries[str], ino1, dirs, files), str ∈ entries, ino ∈ dirs, p0 ∈ tree[str '], str ' ∈ tree, ∀ tree, ino, ino0, D, dirs, ino1, files. ((abs|(tree, ino, p0, ino0) * D))(dirs) ∧ ispath(p0, ino, ino1, dirs, files) ∧ p0 ∈ tree → ino0 = ino1 ⊦ ino0 = ino1

User applied rule constructor cut

And obtained the following new goals

Proof step 8:
dirs[ino] = mkdir(m, md, m0, m1, entries), tree = dnode(md0, sdt), ((abs|(tree, ino, str ' + p0, ino0) * D))(dirs), ispath(p0, entries[str], ino1, dirs, files), str ' ∈ tree, p0 ∈ tree[str '], ino ∈ dirs, str ∈ entries, ∀ tree, ino, ino0, D, dirs, ino1, files. ((abs|(tree, ino, p0, ino0) * D))(dirs) ∧ ispath(p0, ino, ino1, dirs, files) ∧ p0 ∈ tree → ino0 = ino1 ⊦ ino0 = ino1

Rule argument

var: tree

Additional information

Used simplifier rules: directory-tree, entries-tree generated by ∅, dnode, fnode, [ ];;% ⊦ str ' ∈ fnode(fid) ↔ false; (s)