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Proof step 27 from proof subdirs-put (previous step)

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 ⊦ #(subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅)) = 0, subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅) = ∅, str0 = str, 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

Rule argument

pos: right 1

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