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Proof step 9 from proof links-modmeta (previous step)

(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]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
(m1 = ino ⊃ mkdir(m, md, n, n0, entries);dirs[m1]) = mkdir(m2, md1, m3, m4, entries0), dirs[ino] = mkdir(m0, md0, n1, n2, entries), dirs[m1] = mkdir(m5, md2, m6, m7, entries1), ino ∈ dirs, m1 ∈ dirs ⊦ str ∈ entries0 ∧ ino0 = entries0[str] ↔ str ∈ entries1 ∧ ino0 = entries1[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] → [dirs[m1], m5, md2, m6, m7, entries1]

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[m1], m5, md2, m6, m7, entries1