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Proof step 14 from proof links-dirs-union (previous step)

(m ∈ dirs0 ⊃ mkdir(m3, md0, m4, m5, entries0);dirs1[m]) = mkdir(m0, md, m1, m2, entries), dirs0[m] = mkdir(m3, md0, m4, m5, entries0), dirs0 ⊥ dirs1 ⊦ (m ∈ dirs0 ∨ m ∈ dirs1) ∧ str ∈ entries ∧ ino = entries[str] ↔ m ∈ dirs0 ∧ str ∈ entries0 ∧ ino = entries0[str] ∨ m ∈ dirs1 ∧ str ∈ dirs1[m].entries ∧ ino = dirs1[m].entries[str]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
(m ∈ dirs0 ⊃ mkdir(m3, md0, m4, m5, entries0);dirs1[m]) = mkdir(m0, md, m1, m2, entries), dirs0[m] = mkdir(m3, md0, m4, m5, entries0), dirs1[m] = mkdir(m6, md1, m7, m8, entries1), dirs0 ⊥ dirs1 ⊦ (m ∈ dirs0 ∨ m ∈ dirs1) ∧ str ∈ entries ∧ ino = entries[str] ↔ m ∈ dirs0 ∧ str ∈ entries0 ∧ ino = entries0[str] ∨ m ∈ dirs1 ∧ str ∈ entries1 ∧ ino = 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, entries] → [dirs1[m], m4, md1, 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: dirs1[m], m4, md1, m5, entries1