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Proof step 33 from proof subdirs-put (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 34:
dirs[p_ino] = mkdir(m, md, n0, n1, entries), str0 ∈ (subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅)) ∪ (ino ∈ dirs ⊃ ∅ ++ str;∅), p_ino ∈ dirs, ¬ str ∈ entries ∨ ¬ entries[str] ∈ dirs ⊦ # subdirs(p_ino, dirs) = 0, str0 = str, subdirs(p_ino, dirs) = ∅, str0 ∈ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)), ino ∈ dirs

Rule argument

pos: right 4

Additional information

Used simplifier rules: subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;% ⊦ (strings ++ str) ∪ strings0 = (strings ∪ strings0) ++ str; (s)% ⊦ strings ∪ ∅ = strings; (s)%una: ⊦ (strings ∪ strings0) ∪ strings1 = strings ∪ strings0 ∪ strings1; from specification set-union[dir-entries]%unc: ⊦ strings ∪ strings0 = strings0 ∪ strings; from specification set-union[dir-entries]% ⊦ str ≠ str0 → (str ∈ strings ++ str0 ↔ str ∈ strings); (s)% ⊦ str ∈ strings \ strings0 ↔ str ∈ strings ∧ ¬ str ∈ strings0; (ws)