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Proof step 2 from proof abs-unfold (previous step)

dirs[ino] = mkdir(n, md, m0, m1, entries), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), ino ∈ dirs ⊦ abs(tree, ROOT_INO)(dirs) ↔ (∃ sdt. ((abs|(tree, ROOT_INO, p, ino) * ino => mkdir(n, md, m0, m1, entries) * abs(sdt, entries)))(dirs) ∧ p ∈ tree ∧ tree[p] = dnode(md, sdt)), abs(tree0, ROOT_INO)(dirs0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n, md, n0, n1, entries) → # entries = n
Proof step 4:
dirs[ino] = mkdir(# entries, md, # links(ino, dirs), m1, entries), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), ino ∈ dirs ⊦ abs(tree, ROOT_INO)(dirs) ↔ (∃ sdt. ((abs|(tree, ROOT_INO, p, ino) * ino => mkdir(# entries, md, # links(ino, dirs), m1, entries) * abs(sdt, entries)))(dirs) ∧ p ∈ tree ∧ tree[p] = dnode(md, sdt)), abs(tree0, ROOT_INO)(dirs0)

Rule argument

Additional information

Used simplifier rules: make-abs-fold-work: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n, md, n0, n1, entries) → # entries = n;%afs-cons-dir-nlink: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) → # links(ino, dirs) = n; (forward) from specification afs-base