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Proof step 19 from proof subdirs-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
dirs[p_ino] = mkdir(m, md, n0, n1, entries), str0 ∈ (ino ∈ dirs ⊃ subdirs(p_ino, dirs) ++ str;(str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs))), p_ino ∈ dirs, ¬ str ∈ entries ∨ ¬ entries[str] ∈ dirs ⊦ # (ino ∈ dirs ⊃ subdirs(p_ino, dirs) ++ str;(str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs))) = 0, (ino ∈ dirs ⊃ subdirs(p_ino, dirs) ++ str;(str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs))) = ∅, str0 = str, str0 ∈ (ino ∈ dirs ⊃ ∅ ++ str;∅), str0 ∈ entries ∧ entries[str0] ∈ dirs

Rule argument

Additional information

Used simplifier rules: subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)% ⊦ strings \ ∅ = strings; (s)