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Proof step 14 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))) ↔ str0 ∈ (subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅)) ∪ (ino ∈ dirs ⊃ ∅ ++ str;∅), str0 = str

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 15:
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))) ⊦ str0 = str, str0 ∈ (subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅)) ∪ (ino ∈ dirs ⊃ ∅ ++ str;∅)
Proof step 26:
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;∅) ⊦ str0 = str, str0 ∈ (ino ∈ dirs ⊃ subdirs(p_ino, dirs) ++ str;(str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)))

Rule argument

pos: right 1

Proof step 14 from proof subdirs-put (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information