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Proof step 11 from proof subdirs-del-mod-size (previous step)

dirs[p_ino] = mkdir(m, md, n0, n1, entries), str0 ∈ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) -- str ;subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) ), p_ino ∈ dirs, ¬ str ∈ entries ∨ ¬ entries[str] ∈ dirs ⊦ # subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) = 0, subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) = ∅, str0 ∈ subdirs(p_ino, dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
dirs[p_ino] = mkdir(m, md, n0, n1, entries), str0 ∈ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) -- str ;subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) ), p_ino ∈ dirs, ¬ str ∈ entries ∨ ¬ entries[str] ∈ dirs ⊦ # subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) = 0, subdirs(p_ino, dirs[p_ino, mkdir(m0, md, n0, n2, entries)]) = ∅, str0 ∈ entries ∧ entries[str0] ∈ dirs

Rule argument

Additional information

Used simplifier rules: subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)