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Proof step 21 from proof subdirs-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
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

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ strings ≠ ∅ → # strings ≠ 0; (forward) from specification set-basic[dir-entries]%ne-01: ⊦ str ∈ strings → strings ≠ ∅; (forward) from specification set-basic[dir-entries]% ⊦ ¬ str ∈ ∅; (s)