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Proof step 5 from proof subdirs-inc-size (previous step)

p_ino ∈ dirs ⊦ ∀ str. str ∈ subdirs(ino, dirs[p_ino, mkdir(m0 + 1, md, n0, n1, entries)]) ↔ str ∈ subdirs(ino, dirs[p_ino, mkdir(m0, md, n0, n1, entries)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
p_ino ∈ dirs, ino ∈ dirs ⊦ str ∈ (ino = p_ino ⊃ mkdir(m0 + 1, md, n0, n1, entries);dirs[ino]).entries ∧ (ino = p_ino ⊃ mkdir(m0 + 1, md, n0, n1, entries);dirs[ino]).entries[str] ∈ dirs ↔ str ∈ (ino = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[ino]).entries ∧ (ino = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[ino]).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)