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Proof step 5 from proof path-uni (previous step)

ispath(p0, dirs0[n0].entries[name], n1, dirs0, files), n0 ∈ dirs0, name ∈ dirs0[n0].entries, dirs0 ⊥ dirs1, ∀ n0, n1, dirs0, files, dirs1. ispath(p0, n0, n1, dirs0, files) ∧ dirs0 ⊥ dirs1 → ispath(p0, n0, n1, dirs0 ∪ dirs1, files) ⊦ ispath(p0, dirs0[n0].entries[name], n1, dirs0 ∪ dirs1, files)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
dirs0[n0] = mkdir(m, md, m0, m1, entries), ispath(p0, entries[name], n1, dirs0, files), name ∈ entries, n0 ∈ dirs0, dirs0 ⊥ dirs1, ∀ n0, n1, dirs0, files, dirs1. ispath(p0, n0, n1, dirs0, files) ∧ dirs0 ⊥ dirs1 → ispath(p0, n0, n1, dirs0 ∪ dirs1, files) ⊦ ispath(p0, entries[name], n1, dirs0 ∪ dirs1, files)

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] → [dirs0[n0], md, m, entries]

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: dirs0[n0], md, m, entries