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Proof step 18 from proof subdirs-put (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

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;%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)% ⊦ ¬ str ∈ ∅; (s)%size-zero-01: ⊦ strings ≠ ∅ → # strings ≠ 0; (forward) from specification set-basic[dir-entries]%size-zero: ⊦ # strings ≠ 0 → strings ≠ ∅; (forward) from specification set-basic[dir-entries]% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)% ⊦ ∅ ≠ strings ++ str; (s)% ⊦ str ≠ str0 → (str ∈ strings ++ str0 ↔ str ∈ strings); (s)% ⊦ str ∈ ∅ ++ str0 ↔ str = str0; (s)