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Proof step 5 from proof subdirs-modmeta (previous step)

dirs[ino] = mkdir(m0, md0, n1, n2, entries), ino ∈ dirs ⊦ ∀ str. str ∈ subdirs(ino0, dirs[ino, mkdir(m, md, n, n0, entries)]) ↔ str ∈ subdirs(ino0, dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
dirs[ino] = mkdir(m0, md0, n1, n2, entries), ino ∈ dirs, ino0 ∈ dirs ⊦ str ∈ (ino0 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[ino0]).entries ∧ (ino0 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[ino0]).entries[str] ∈ dirs ↔ str ∈ dirs[ino0].entries ∧ dirs[ino0].entries[str] ∈ dirs

Rule argument

Additional information

Used simplifier rules: subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;% ⊦ dirs[ino, ad][ino0] = (ino0 = ino ∧ ino ∈ dirs ⊃ ad;dirs[ino0]); (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)