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Proof step 148 from proof afs-cons-rename-dir-negdentry (previous step)

(m0 = m ⊃ mkdir(m4 + 1, md0, m6, m7 + 1, entries0[str0, entries2[str]]);mkdir(m4 + 1, md3, m6, m7 + 1, entries0[str0, entries2[str]])).meta:= md1 = mkdir(m15, md7, m16, m17, entries4), (m0 = m ⊃ mkdir(m1, md, # links(m, dirs0), m2, entries2 -- str) ;mkdir(# entries, md6, # links(m0, dirs0), # subdirs(m0, dirs0), entries) ) = mkdir(m4, md3, m6, m7, entries0), dirs0[entries2[str]] = mkdir(m3, md4, # links(entries2[str], dirs0), m5, entries1), dirs0[m0] = mkdir(# entries, md6, # links(m0, dirs0), # subdirs(m0, dirs0), entries), dirs0[m] = mkdir(m1 + 1, md, # links(m, dirs0), m2 + 1, entries2), files[m0] = mkfile(m9, md2, m11, content), m0 = m, afs-cons(dirs0, files), m0 ∈ dirs0, ROOT_INO ∈ dirs0, entries2[str] ∈ dirs0, str ∈ entries2, m ∈ dirs0, # links(m0, dirs0) ≤ 1, m0 = m → md1 = md0, name ∈ entries → entries[name] ∈ dirs0 ∨ entries[name] ∈ files, m0 ≠ ROOT_INO ∧ links(m0, dirs0) = ∅ → entries = ∅, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ m15 = # entries4, str0 ∈ entries

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 149:
# subdirs(m, dirs0) = m2 + 1, # entries2 = m4 + 1, dirs0[m] = mkdir(m4 + 1, md, # links(m, dirs0), m2 + 1, entries2), dirs0[entries2[str]] = mkdir(m3, md4, # links(entries2[str], dirs0), m5, entries1), files[m] = mkfile(m9, md2, m11, content), afs-cons(dirs0, files), str ∈ entries2, entries2[str] ∈ dirs0, ROOT_INO ∈ dirs0, m ∈ dirs0, # links(m, dirs0) ≤ 1, name ∈ entries2 → entries2[name] ∈ dirs0 ∨ entries2[name] ∈ files, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ m4 + 1 = # (entries2 -- str)[str0, entries2[str]], entries2 = ∅, str0 ∈ entries2, m ≠ ROOT_INO ∧ links(m, dirs0) = ∅

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ # ∅ = 0; (s)%⊦ mkdir(m, md, m0, m1, entries).meta:= md0 = mkdir(m, md0, m0, m1, entries); (s)%⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)