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Proof step 104 from proof inv-evict (previous step)

# links(m, dirs) = 0, dirs[m] = mkdir(m0, md, 0, m2, entries), dirs[m3] = mkdir(m1, md0, # links(m3, dirs), m4, entries0), dcache[dentrykey(m3, str)] = mkdentry(str, entries0[str]), (dirs -- m)[m3] = mkdir(m5, md1, 0, m7, entries1), afs-cons(dirs, files), afs-cons(dirs -- m, files), synced(dirs, icache, pcache), str ∈ entries0, m ∈ dirs, dentrykey(m3, str) ∈ dcache, m3 ∈ dirs, asynced1 → synced(files, icache, pcache), ∀ key0. key0 ∈ dcache → key0.dentry? ∧ key0.name = dcache[key0].name ∧ key0.ino ∈ dirs ∧ (links(key0.ino, dirs) ≠ ∅ ∨ key0.ino = ROOT_INO) ∧ (dcache[key0].ino ∈ dirs ∨ dcache[key0].ino ∈ files) ∧ dcache[key0].name ∈ dirs[key0.ino].entries ∧ dcache[key0].ino = dirs[key0.ino].entries[dcache[key0].name] ∧ dcache[key0].dentry?, entries0[str] ∈ dirs ∨ entries0[str] ∈ files ⊦ # links(m3, dirs) = 0, links(m3, dirs) = ∅, m = ROOT_INO, m3 = ROOT_INO, (links(m3, dirs -- m) ≠ ∅ ∨ m3 = ROOT_INO) ∧ (entries0[str] ≠ m ∧ entries0[str] ∈ dirs ∨ entries0[str] ∈ files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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