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Proof step 22 from proof subdirs-put (previous step)

dirs[p_ino] = mkdir(m, md, n0, n1, entries), p_ino ∈ dirs, str0 ∈ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)), ¬ str ∈ entries ∨ ¬ entries[str] ∈ dirs ⊦ # (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)) = 0, (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)) = ∅, str0 = str, ino ∈ dirs, str0 ∈ entries ∧ entries[str0] ∈ dirs

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 23:
dirs[p_ino] = mkdir(m, md, n0, n1, entries), str0 ∈ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)), p_ino ∈ dirs, ¬ str ∈ entries ∨ ¬ entries[str] ∈ dirs ⊦ # subdirs(p_ino, dirs) = 0, (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)) = ∅, str0 = str, ino ∈ dirs, str0 ∈ entries ∧ entries[str0] ∈ dirs

Rule argument

pos: right 1

Additional information

Used simplifier rules: subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;% ⊦ strings -- str = ∅ ↔ strings = ∅ ∨ strings = ∅ ++ str; (ws)% ⊦ str ∈ strings -- str0 ↔ str ≠ str0 ∧ str ∈ strings; (s)