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Proof step 8 from proof subdirs-alloc (previous step)

(dirs ++ ino)[p_ino] = mkdir(m, md, m0, m1, entries), dirs[p_ino] = mkdir(m2, md0, m3, m4, entries0), p_ino ∈ dirs ⊦ str ∈ entries ∧ (entries[str] = ino ∨ entries[str] ∈ dirs) ↔ str ∈ entries0 ∧ entries0[str] ∈ dirs, ino = 0, ino ∈ dirs, ino ∈ entries0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: p_ino = ino

Additional information

Used simplifier rules: ino ∈ dirs, ino0 ≠ 0, ¬ ino0 ∈ dirs ⊦ ino ≠ ino0 → (dirs ++ ino0)[ino] = dirs[ino]; (s)%⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)%str ∈ entries ⊦ entries[str] ∈ entries; (s) Used formulas: p_ino = ino