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Proof step 115 from proof inv-unlink (previous step)

# links(entries0[str], dirs0) = m2 + 1, icache0[entries0[str], mkientry(false, mkinode(entries0[str], md0, false, m2, 0, m3))][ino] = mkientry(true, inode0), icache0[ino] = mkientry(false, inode), dirs0[m0] = mkdir(m8 + 1, md2, # links(m0, dirs0), m7, entries0), files0[entries0[str]] = mkfile(m1, md1, m2 + 1, content0), pcache[datakey(ino, n)] = mkpentry(boolvar, buf), afs-cons(dirs0, files0), afs-cons(dirs0[m0, mkdir(m8, md3, # links(m0, dirs0), m7, entries0 -- str)], files0[entries0[str], mkfile(m3, md0, m2, content0)]), synced(files0, icache0, pcache), ino ∈ icache0, ino ∈ files0, str ∈ entries0, entries0[str] ∈ files0, m0 ∈ dirs0, datakey(ino, n) ∈ pcache → ¬ boolvar, ∀ n. datakey(ino, n) ∈ pcache → ¬ pcache[datakey(ino, n)].dirty, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ ino = m0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: ino = entries0[str]

Additional information

Used simplifier rules: ⊦ ino ≠ ino0 → icache[ino0, ie][ino] = icache[ino]; (s)%⊦ mkientry(boolvar, inode) = mkientry(boolvar0, inode0) ↔ (boolvar ↔ boolvar0) ∧ inode = inode0; (ws)% ⊦ icache[ino, ie][ino] = ie; (s) Used formulas: ino = entries0[str]