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Proof step 20 from proof links-del-entry (previous step)

ino0 ∈ dirs, str ∈ entries, m0 ∈ dirs ⊦ str0 ∈ (m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries -- str);dirs[m0]).entries ∧ ino = (m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries -- str);dirs[m0]).entries[str0] ↔ str0 ∈ (m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries);dirs[m0]).entries ∧ ino = (m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries);dirs[m0]).entries[str0], entries[str] = ino

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
(m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries -- str);dirs[m0]) = mkdir(m1, md0, m2, m3, entries0), m0 ∈ dirs, str ∈ entries, ino0 ∈ dirs ⊦ str0 ∈ entries0 ∧ ino = entries0[str0] ↔ str0 ∈ (m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries);dirs[m0]).entries ∧ ino = (m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries);dirs[m0]).entries[str0], entries[str] = ino

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] → [(m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries -- str);dirs[m0]), m1, md0, 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: (m0 = ino0 ⊃ mkdir(m, md, n0, n1, entries -- str);dirs[m0]), m1, md0, m2, m3, entries0