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Proof step 108 from proof afs-cons-rename-dir-negdentry (previous step)

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

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: entries2[str] = 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) Used formulas: entries2[str] = m0