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Proof step 23 from proof subdirs-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
dirs[p_ino] = mkdir(m, md, n0, n1, entries), str0 ∈ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)), p_ino ∈ dirs, ¬ str ∈ entries ∨ ¬ entries[str] ∈ dirs ⊦ # subdirs(p_ino, dirs) = 0, (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs)) = ∅, str0 = str, subdirs(p_ino, dirs) = ∅, 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]%size-zero: ⊦ # strings ≠ 0 → strings ≠ ∅; (forward) from specification set-basic[dir-entries]