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Proof step 6 from proof absc-path (previous step)

((abs|(tree, ino, str ' + p0, ino0) * D))(dirs), ispath(p0, dirs[ino].entries[str], ino1, dirs, files), str ' ∈ tree, p0 ∈ tree[str '], ino ∈ dirs, str ∈ dirs[ino].entries, ∀ tree, ino, ino0, D, dirs, ino1, files. ((abs|(tree, ino, p0, ino0) * D))(dirs) ∧ ispath(p0, ino, ino1, dirs, files) ∧ p0 ∈ tree → ino0 = ino1 ⊦ ino0 = ino1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
dirs[ino] = mkdir(m, md, m0, m1, entries), ((abs|(tree, ino, str ' + p0, ino0) * D))(dirs), ispath(p0, entries[str], ino1, dirs, files), str ∈ entries, ino ∈ dirs, p0 ∈ tree[str '], str ' ∈ tree, ∀ tree, ino, ino0, D, dirs, ino1, files. ((abs|(tree, ino, p0, ino0) * D))(dirs) ∧ ispath(p0, ino, ino1, dirs, files) ∧ p0 ∈ tree → ino0 = ino1 ⊦ ino0 = ino1

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] → [dirs[ino], m, md, m0, 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: dirs[ino], m, md, m0, entries