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Proof step 5 from proof abs-unfold (previous step)

dirs[ino] = mkdir(# entries, md, # links(ino, dirs), m1, entries), abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), ino ∈ dirs ⊦ abs(tree0, ROOT_INO)(dirs0), ∃ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files))
Proof step 7:
⊦ ispath(p, ino, m, dirs, files) ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → p ∈ tree ∧ (tree[p].dir? ↔ m ∈ dirs)
Proof step 8:
dirs[ino] = mkdir(# entries, md, # links(ino, dirs), m1, entries), tree[p].dir?, abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), ino ∈ dirs ⊦ abs(tree0, ROOT_INO)(dirs0), ∃ sdt. ((abs|(tree, ROOT_INO, p, ino) * ino => mkdir(# entries, md, # links(ino, dirs), m1, entries) * abs(sdt, entries)))(dirs) ∧ tree[p] = dnode(md, sdt)

Rule argument

Additional information

Used simplifier rules: path-in-eq: ⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files));%path-in: ⊦ ispath(p, ino, m, dirs, files) ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → p ∈ tree ∧ (tree[p].dir? ↔ m ∈ dirs);% ⊦ ispath(p, n, m, dirs, files) → ispath(p, n, dirs, files); (s)