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Proof step 262 from proof inv-write_inode (previous step)

icache[ino] = mkientry(false, inode), icache[m] = mkientry(boolvar, mkinode(m0, md0, boolvar0, m1, m3, m4)), pcache[datakey(ino, n)] = mkpentry(boolvar1, buf), icache[m0, mkientry(false, mkinode(m0, md, boolvar0, m1, m3, m2))][ino] = mkientry(true, inode0), afs-cons(dirs, files), afs-cons(dirs, files[m, (files[m].meta:= md).size:= m2]), synced(files, icache, pcache), ino ∈ icache, ino ∈ icache[m0, mkientry(false, mkinode(m0, md, boolvar0, m1, m3, m2))], ino ∈ files, m ∈ icache, m ∈ files, datakey(ino, n) ∈ pcache → ¬ boolvar1, ∀ n. datakey(ino, n) ∈ pcache → ¬ pcache[datakey(ino, n)].dirty ⊦

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: ino = m0

Additional information

Used simplifier rules: ⊦ ino ≠ ino0 → icache[ino0, ie][ino] = icache[ino]; (s)% ⊦ ino ≠ ino0 → (ino ∈ icache[ino0, ie] ↔ ino ∈ icache); (s)%⊦ mkientry(boolvar, inode) = mkientry(boolvar0, inode0) ↔ (boolvar ↔ boolvar0) ∧ inode = inode0; (ws)% ⊦ icache[ino, ie][ino] = ie; (s) Used formulas: ino = m0