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Proof step 29 from proof subdirs-put (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 6

Additional information

Used simplifier rules: subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;%subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;% ⊦ 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 ∈ strings -- str0 ↔ str ≠ str0 ∧ str ∈ strings; (s)% ⊦ str ≠ str0 → (str ∈ strings -- str0 ↔ str ∈ strings); (ss)% ⊦ (strings ++ str) ∪ strings0 = (strings ∪ strings0) ++ str; (s)% ⊦ (strings -- str) ++ str = strings ++ str; (s)% ⊦ str ≠ str0 → (str ∈ strings ++ str0 ↔ str ∈ strings); (s)