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Proof step 152 from proof afs-cons-rename-dir-negdentry (previous step)

(ino = m ⊃ mkdir(m1, md0, # links(m, dirs0), m2, entries2 -- str) ;mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries) ) = mkdir(m15, md7, m16, m17, entries4), dirs0[ino] = mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries), files[ino] = mkfile(m9, md2, m11, content), dirs0[m] = mkdir(m1 + 1, md, # links(m, dirs0), m2 + 1, entries2), dirs0[m0] = mkdir(m8, md5, # links(m0, dirs0), m10, entries3), dirs0[entries2[str]] = mkdir(m3, md4, # links(entries2[str], dirs0), m5, entries1), afs-cons(dirs0, files), m ∈ dirs0, str ∈ entries2, entries2[str] ∈ dirs0, m0 ∈ dirs0, ROOT_INO ∈ dirs0, ino ∈ dirs0, # links(ino, dirs0) ≤ 1, ino ≠ ROOT_INO ∧ links(ino, dirs0) = ∅ → entries = ∅, name ∈ entries → entries[name] ∈ dirs0 ∨ entries[name] ∈ files, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ ino = m0, m0 = m, m15 = # entries4, str0 ∈ entries3

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)% ⊦ str ∈ entries → #(entries -- str) = (# entries)-1; (s)% ⊦ n -1 = n - 1; (s)% ⊦ # ∅ = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s) Used formulas: ino = m