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Proof step 152 from proof inv-mkdir (previous step)

dirs0[m0, mkdir(m3 + 1, md1, # links(m0, dirs0), m5 + 1, entries0[str, ino0])][m0] = mkdir(m, md0, m1, m2, entries), dirs0[m0] = mkdir(m3, md1, # links(m0, dirs0), m5, entries0), afs-cons ((dirs0[m0, mkdir(m3 + 1, md1, # links(m0, dirs0), m5 + 1, entries0[str, ino0])] ++ ino0)[ino0, mkdir(0, md, 1, 0, ∅)] [m0, mkdir(m, md2, m1, m2, entries) ], files), afs-cons(dirs0, files), dcache-cons(dcache, dirs0, files), synced(dirs0, icache0, pcache), m0 ∈ dirs0, m0 ∈ dirs0[m0, mkdir(m3 + 1, md1, # links(m0, dirs0), m5 + 1, entries0[str, ino0])], asynced1 → synced(files, icache0, pcache), links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ ino0 = 0, m0 = ino0, dcache-cons (dcache[dentrykey(m0, str), mkdentry(str, ino0)], (dirs0[m0, mkdir(m3 + 1, md1, # links(m0, dirs0), m5 + 1, entries0[str, ino0])] ++ ino0)[ino0, mkdir(0, md, 1, 0, ∅)] [m0, mkdir(m, md2, m1, m2, entries) ], files), str ∈ entries0, ino0 ∈ files, ino0 ∈ dirs0, ino0 ∈ dirs0[m0, mkdir(m3 + 1, md1, # links(m0, dirs0), m5 + 1, entries0[str, ino0])]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 153:
dirs0[m0] = mkdir(m3, md1, # links(m0, dirs0), m5, entries0), afs-cons ((dirs0[m0, mkdir(m3 + 1, md1, # links(m0, dirs0), m5 + 1, entries0[str, ino0])] ++ ino0)[ino0, mkdir(0, md, 1, 0, ∅)] [m0, mkdir(m3 + 1, md2, # links(m0, dirs0), m5 + 1, entries0[str, ino0]) ], files), afs-cons(dirs0, files), dcache-cons(dcache, dirs0, files), synced(dirs0, icache0, pcache), m0 ∈ dirs0, asynced1 → synced(files, icache0, pcache), links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ ino0 = 0, dcache-cons (dcache[dentrykey(m0, str), mkdentry(str, ino0)], (dirs0[m0, mkdir(m3 + 1, md1, # links(m0, dirs0), m5 + 1, entries0[str, ino0])] ++ ino0)[ino0, mkdir(0, md, 1, 0, ∅)] [m0, mkdir(m3 + 1, md2, # links(m0, dirs0), m5 + 1, entries0[str, ino0]) ], files), str ∈ entries0, ino0 ∈ files, ino0 ∈ dirs0

Rule argument

Additional information

Used simplifier rules: ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)%⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)% ⊦ dirs[ino, ad][ino0] = (ino0 = ino ∧ ino ∈ dirs ⊃ ad;dirs[ino0]); (s)