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Proof step 7 from proof subdirs-inc-size (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
(ino = p_ino ⊃ mkdir(m0 + 1, md, n0, n1, entries);dirs[ino]) = mkdir(m, md0, m1, m2, entries0), (ino = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[ino]) = mkdir(m3, md1, m4, m5, entries1), p_ino ∈ dirs, ino ∈ dirs ⊦ str ∈ entries0 ∧ entries0[str] ∈ dirs ↔ str ∈ entries1 ∧ entries1[str] ∈ dirs

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] → [(ino = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[ino]), m3, md1, m4, m5, 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: (ino = p_ino ⊃ mkdir(m0, md, n0, n1, entries);dirs[ino]), m3, md1, m4, m5, entries1