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Proof step 4 from proof abs-unfold (previous step)

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)

User applied rule case distinction

And obtained the following new goals

Proof step 5:
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)
Proof step 18:
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, ∃ 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), abs(tree, ROOT_INO)(dirs)

Rule argument

pos: right 1

Proof step 4 from proof abs-unfold (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information