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Proof step 155 from proof afs-cons-rename-dir-negdentry (previous step)

(ino = m0 ⊃ mkdir(m4 + 1, md1, # links(m0, dirs0), m7 + 1, (entries2 -- str)[str0, entries2[str]]) ;mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries) ) = mkdir(m15, md7, m16, m17, entries4), dirs0[m0] = mkdir(m4 + 1, md, # links(m0, dirs0), m7 + 1, entries2), dirs0[entries2[str]] = mkdir(m3, md4, # links(entries2[str], dirs0), m5, entries1), dirs0[ino] = mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries), files[ino] = mkfile(m9, md2, m11, content), afs-cons(dirs0, files), 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(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ m17 = # subdirs(ino, dirs0[m0, mkdir(m4, md1, # links(m0, dirs0), m7 + 1, (entries2 -- str)[str0, entries2[str]])]), str0 ∈ entries2

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 156:
(ino = m0 ⊃ mkdir(m4 + 1, md1, # links(m0, dirs0), m7 + 1, (entries2 -- str)[str0, entries2[str]]) ;mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries) ) = mkdir(m15, md7, m16, m17, entries4), files[ino] = mkfile(m9, md2, m11, content), dirs0[ino] = mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries), dirs0[entries2[str]] = mkdir(m3, md4, # links(entries2[str], dirs0), m5, entries1), dirs0[m0] = mkdir(m4 + 1, md, # links(m0, dirs0), m7 + 1, entries2), ino = m0, afs-cons(dirs0, files), ino ∈ dirs0, ROOT_INO ∈ dirs0, m0 ∈ dirs0, entries2[str] ∈ dirs0, str ∈ entries2, # links(ino, dirs0) ≤ 1, name ∈ entries → entries[name] ∈ dirs0 ∨ entries[name] ∈ files, ino ≠ ROOT_INO ∧ links(ino, dirs0) = ∅ → entries = ∅, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ m17 = # subdirs(ino, dirs0[m0, mkdir(m4, md1, # links(m0, dirs0), m7 + 1, (entries2 -- str)[str0, entries2[str]])]), str0 ∈ entries2

Rule argument

fma: ino = m0

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)% ⊦ ino0 ≠ ino1 → subdirs(ino0, dirs[ino1, ad]) = subdirs(ino0, dirs); (s) Used formulas: ino = m0