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Proof step 8 from proof links-modmeta (previous step)

dirs[ino] = mkdir(m0, md0, n1, n2, entries), ino ∈ dirs, m1 ∈ dirs ⊦ str ∈ (m1 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[m1]).entries ∧ ino0 = (m1 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[m1]).entries[str] ↔ str ∈ dirs[m1].entries ∧ ino0 = dirs[m1].entries[str]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
(m1 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[m1]) = mkdir(m2, md1, m3, m4, entries0), dirs[ino] = mkdir(m0, md0, n1, n2, entries), m1 ∈ dirs, ino ∈ dirs ⊦ str ∈ entries0 ∧ ino0 = entries0[str] ↔ str ∈ dirs[m1].entries ∧ ino0 = dirs[m1].entries[str]

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] → [(m1 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[m1]), m2, md1, m3, m4, 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: (m1 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[m1]), m2, md1, m3, m4, entries0