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Proof step 8 from proof links-inc-nsubdirs (previous step)

p_ino ∈ dirs, m ∈ dirs ⊦ str ∈ (m = p_ino ⊃ mkdir(m0, md, n0, n1 + 1, entries);dirs[m]).entries ∧ ino = (m = p_ino ⊃ mkdir(m0, md, n0, n1 + 1, entries);dirs[m]).entries[str] ↔ str ∈ (m = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[m]).entries ∧ ino = (m = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[m]).entries[str]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
(m = p_ino ⊃ mkdir(m0, md, n0, n1 + 1, entries);dirs[m]) = mkdir(m1, md0, m2, m3, entries0), m ∈ dirs, p_ino ∈ dirs ⊦ str ∈ entries0 ∧ ino = entries0[str] ↔ str ∈ (m = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[m]).entries ∧ ino = (m = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[m]).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] → [(m = p_ino ⊃ mkdir(m0, md, n0, n1 + 1, entries);dirs[m]), 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: (m = p_ino ⊃ mkdir(m0, md, n0, n1 + 1, entries);dirs[m]), m1, md0, m2, m3, entries0