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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
(ino0 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[ino0]) = mkdir(m1, md1, m2, m3, entries0), dirs[ino] = mkdir(m0, md0, n1, n2, entries), ino0 ∈ dirs, ino ∈ dirs ⊦ str ∈ entries0 ∧ entries0[str] ∈ dirs ↔ str ∈ dirs[ino0].entries ∧ dirs[ino0].entries[str] ∈ 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] → [(ino0 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[ino0]), m1, md1, m2, m3, entries0]

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: (ino0 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[ino0]), m1, md1, m2, m3, entries0