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Proof step 108 from proof afs-cons-rename-file-dentry (previous step)

# links(ino, dirs0) = 0, # links(entries[str], dirs0) = m11 + 1, (ino = m ⊃ mkdir(m6, md1, # links(m, dirs0), m10, entries[str, entries1[str0]]) ;(ino = m0 ⊃ mkdir(m3, md2, # links(m0, dirs0), m7, entries1 -- str0);mkdir(0, md9, 0, # subdirs(ino, dirs0), ∅)) ) = mkdir(m19, md10, m20, m21, entries3), files0[ino] = mkfile(m12, md7, m13, content0), dirs0[m] = mkdir(m6, md6, # links(m, dirs0), m10, entries), files0[entries[str]] = mkfile(m5, md0, m11 + 1, content1), files0[entries1[str0]] = mkfile(m1, md3, # links(entries1[str0], dirs0), content), dirs0[m0] = mkdir(m3 + 1, md4, # links(m0, dirs0), m7, entries1), dirs0[ino] = mkdir(0, md9, 0, # subdirs(ino, dirs0), ∅), afs-cons(dirs0, files0), ino ∈ dirs0, m ∈ dirs0, str0 ∈ entries1, entries1[str0] ∈ files0, str ∈ entries, m0 ∈ dirs0, entries[str] ∈ files0, ROOT_INO ∈ dirs0, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ entries3 = ∅, ino = ROOT_INO, m = m0, entries1[str0] ∈ dirs0, entries[str] ∈ dirs0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: ino = m

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-01: ⊦ ls ≠ ∅ → # ls ≠ 0; (forward) from specification set-basic[alink-set]%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 Used formulas: ino = m