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Proof step 110 from proof inv-rmdir (previous step)

(entries0[str] = m ⊃ mkdir(m3, md3, # links(entries0[str], dirs0), m2 - 1, entries0 -- str) ;mkdir(m1, md2, # links(entries0[str], dirs0), m4, ∅) ) = mkdir(m0, md, 0, m6, entries), dirs0[m] = mkdir(m3 + 1, md3, # links(m, dirs0), m2, entries0), dirs0[entries0[str]] = mkdir(m1, md2, # links(entries0[str], dirs0), m4, ∅), afs-cons(dirs0, files), m ∈ dirs0, str ∈ entries0, entries0[str] ∈ dirs0, 1 ≤ m2 ⊦ # links(m, dirs0) = 0, entries0[str] = ROOT_INO, links(m, dirs0) = ∅, m = ROOT_INO, m2 = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)%size-zero-02: ⊦ # ls = 0 → ls = ∅; (forward) from specification set-basic[alink-set]%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%⊦ p_ino ∈ dirs ∧ dirs[p_ino] = mkdir(m, md, n, n0, entries) ∧ str ∈ entries → links(entries[str], dirs) ≠ ∅; (s)