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

p_ino ∈ dirs ⊦ ino = 0, ino ∈ dirs[p_ino].entries, ino ∈ dirs, ∀ str. str ∈ subdirs(p_ino, dirs ++ ino) ↔ str ∈ subdirs(p_ino, dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
p_ino ∈ dirs ⊦ str ∈ (dirs ++ ino)[p_ino].entries ∧ ((dirs ++ ino)[p_ino].entries[str] = ino ∨ (dirs ++ ino)[p_ino].entries[str] ∈ dirs) ↔ str ∈ dirs[p_ino].entries ∧ dirs[p_ino].entries[str] ∈ dirs, ino = 0, ino ∈ dirs, ino ∈ dirs[p_ino].entries

Rule argument

Additional information

Used simplifier rules: subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;%ino ≠ 0, ¬ ino ∈ dirs ⊦ ino0 ∈ dirs ++ ino ↔ ino0 = ino ∨ ino0 ∈ dirs; (s)