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Proof step 158 from proof inv-link (previous step)

dirs[ino0] = mkdir(m0, md0, # links(ino0, dirs), m3, entries0), files[entries0[str]] = mkfile(m, md, # links(entries0[str], dirs), content), dirs[ino1] = mkdir(m1, md1, # links(ino1, dirs), m4, entries), afs-cons(dirs, files), ispath(parent(p), ROOT_INO, ino1, dirs, files), ispath(q, ROOT_INO, ino0, dirs, files), of-cons(of, dirs, files), pw(user, md1), px(user, md1), px(user, md0), px-parent(user, parent(p), ROOT_INO, dirs, files), px-parent(user, q, ROOT_INO, dirs, files), ino1 ∈ dirs, ino0 ∈ dirs, entries0[str] ∈ files, str ∈ entries0, links(ino1, dirs) ≠ ∅ ∨ ino1 = ROOT_INO, links(ino0, dirs) ≠ ∅ ∨ ino0 = ROOT_INO ⊦ p = ε, entries0[str] ∈ dirs, p.last ∈ entries, ∃ ino. ino ∈ dirs ∧ (links(ino, dirs) ≠ ∅ ∨ ino = ROOT_INO) ∧ str ∈ dirs[ino].entries ∧ entries0[str] = dirs[ino].entries[str]

User applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: ino0 abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s) Used terms: ino0