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Proof step 9 from proof link-mod-dir-meta (previous step)

(m0 = m ⊃ dirs[m].meta:= md;dirs[m0]) = mkdir(m1, md0, m2, m3, entries), m0 ∈ dirs, m ∈ dirs ⊦ str ∈ entries ∧ ino = entries[str] ↔ str ∈ dirs[m0].entries ∧ ino = dirs[m0].entries[str]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
(m0 = m ⊃ dirs[m].meta:= md;dirs[m0]) = mkdir(m1, md0, m2, m3, entries), dirs[m0] = mkdir(m4, md1, m5, m6, entries0), m ∈ dirs, m0 ∈ dirs ⊦ str ∈ entries ∧ ino = entries[str] ↔ str ∈ entries0 ∧ ino = entries0[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, md, m, entries] → [dirs[m0], md1, m2, 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: dirs[m0], md1, m2, entries0