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Proof step 6 from proof subdirs-alloc (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
(dirs ++ ino)[p_ino] = mkdir(m, md, m0, m1, entries), p_ino ∈ dirs ⊦ str ∈ entries ∧ (entries[str] = ino ∨ entries[str] ∈ dirs) ↔ str ∈ dirs[p_ino].entries ∧ dirs[p_ino].entries[str] ∈ dirs, ino = 0, ino ∈ dirs[p_ino].entries, ino ∈ dirs

Rule argument

spec: abstract-dir name: elim seq: ⊦ ( ((m = ad.size ∧ md = ad.meta) ∧ m0 = ad.nlink) ∧ m1 = ad.nsubdirs) ∧ entries = ad.entries ↔ ad = mkdir(m, md, m0, m1, entries) subst: [ad, m, md, m0, m1, entries] → [(dirs ++ ino)[p_ino], m, md, m0, m1, entries]

Additional information

Used simplifier rules: ⊦ ( ((m = ad.size ∧ md = ad.meta) ∧ m0 = ad.nlink) ∧ m1 = ad.nsubdirs) ∧ entries = ad.entries ↔ ad = mkdir(m, md, m0, m1, entries); (elim) Used terms: (dirs ++ ino)[p_ino], m, md, m0, m1, entries