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Proof step 133 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), str ∈ entries2, m0 ∈ dirs0, ROOT_INO ∈ dirs0, entries2[str] ∈ dirs0, mklink(m0, str) ∈ links(entries2[str], dirs0), # links(entries2[str], dirs0) ≤ 1, name ∈ entries1 → entries1[name] ∈ dirs0 ∨ entries1[name] ∈ files, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ #(links(entries2[str], dirs0) -- mklink(m0, str)) = 0, links(entries2[str], dirs0) = ∅ ++ mklink(m0, str), str0 ∈ entries2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)% ⊦ mklink(m, str).name = str; (s)% ⊦ mklink(m, str).pino = m; (s)%⊦ alink ∈ links(ino, dirs) ↔ alink.pino ∈ dirs ∧ alink.name ∈ dirs[alink.pino].entries ∧ ino = dirs[alink.pino].entries[alink.name]; (s)% ⊦ alink ∈ ls → #(ls -- alink) = # ls - 1; (s)