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Proof step 136 from proof afs-cons-rename-dir-dentry (previous step)

# links(entries2[str], dirs0) = m12 + 1, (entries2[str0] = m ⊃ mkdir(m8, md1, # links(entries2[str0], dirs0), m11, (entries2 -- str0)[str, entries2[str0]]) ;mkdir(# entries1, md3, # links(entries2[str0], dirs0), # subdirs(entries2[str0], dirs0), entries1) ) = mkdir(m20, md10, m21, m22, entries5), files[entries2[str0]] = mkfile(m14, md8, m16, content), dirs0[entries2[str0]] = mkdir(# entries1, md3, # links(entries2[str0], dirs0), # subdirs(entries2[str0], dirs0), entries1), dirs0[m] = mkdir(m8 + 1, md0, # links(m, dirs0), m11 + 1, entries2), dirs0[entries2[str]] = mkdir(m13, md6, m12 + 1, m15, ∅), afs-cons(dirs0, files), ROOT_INO ∈ dirs0, entries2[str] ∈ dirs0, str ∈ entries2, m ∈ dirs0, entries2[str0] ∈ dirs0, str0 ∈ entries2, # links(entries2[str0], dirs0) ≤ 1, name ∈ entries1 → entries1[name] ∈ dirs0 ∨ entries1[name] ∈ files, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ # links(entries2[str0], dirs0) = m21, entries2[str0] = entries2[str], entries2[str] = ROOT_INO, entries2[str] = m, mkdentry(str, entries2[str]) = mkdentry(str0, entries2[str0])

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: entries2[str0] = 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) Used formulas: entries2[str0] = m