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Proof step 81 from proof inv-unlink (previous step)

# links(entries0[str], dirs0) = m1 + 1, files0[entries0[str]] = mkfile(m0, md1, m1 + 1, content), dirs0[m] = mkdir(m2 + 1, md2, # links(m, dirs0), m3, entries0), dirs = dirs0[m, mkdir(m2, md, # links(m, dirs0), m3, entries0 -- str)], files = files0[entries0[str], mkfile(m0, md1, m1, content)], ino = inode.ino, inode = mkinode(entries0[str], md0, false, # links(entries0[str], dirs0), 0, m6), afs-cons(dirs0[m, mkdir(m2, md2, # links(m, dirs0), m3, entries0 -- str)], files0[entries0[str], mkfile(m0, md1, m1, content)]), afs-cons(dirs0, files0), afs-cons(dirs0[m, mkdir(m2, md, # links(m, dirs0), m3, entries0 -- str)], files0[entries0[str], mkfile(m0, md1, m1, content)]), afs-cons(dirs, files[ino, files[ino].size:= inode.size][ino, files[ino, files[ino].size:= inode.size][ino].meta:= inode.meta]), m ∈ dirs0, str ∈ entries0, entries0[str] ∈ files0, ino ∈ files, (entries0[str] ∈ dirs ⊃ str ∈ entries0 ∧ str ∈ entries0 ∧ str ∈ entries0 ∧ str ∈ entries0 ∧ str ∈ entries0 ;str ∈ entries0 ∧ str ∈ entries0 ∧ str ∈ entries0 ∧ str ∈ entries0 ), links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ ⟪asynced := ?⟫ true

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

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