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Proof step 98 from proof inv-mkdir (previous step)

icache0[ino] = mkientry(false, inode), dirs0[m0] = mkdir(m3, md1, # links(m0, dirs0), m5, entries0), pcache[datakey(ino, n)] = mkpentry(boolvar, buf), icache0[m0, mkientry(false, mkinode(m0, md2, true, # links(m0, dirs0), m5 + 1, m3 + 1))] [ino0, mkientry(false, mkinode(ino0, md, true, 1, 0, 0)) ] [ino ] = mkientry(true, inode0), afs-cons(dirs0, files), 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), synced(files, icache0, pcache), ino ∈ icache0, m0 ∈ dirs0, ino ∈ files, datakey(ino, n) ∈ pcache → ¬ boolvar, ∀ n. datakey(ino, n) ∈ pcache → ¬ pcache[datakey(ino, n)].dirty, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ ino = m0, ino0 = 0, ino0 ∈ dirs0, ino0 ∈ files, str ∈ entries0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: ino = ino0

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) Used formulas: ino = ino0