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Proof step 24 from proof abs-has-rootpath-h (previous step)

((ino0 => mkdir(n0, md, m0, n, entries0)))(dirs2), ino ∈ dirs2, dirs2 ⊥ dirs1, ∀ sdt0, ino0, n0, md, m0, n, entries0, entries, dirs0, dirs, ino, files. # sdt0 < # ∅ → (((ino0 => mkdir(n0, md, m0, n, entries0) * abs(sdt0, entries)) * {dirs0} ))(dirs) ∧ ino ∈ dirs ∧ entries ⊆ entries0 ∧ (∀ str. str ∈ sdt0 → (∀ ino, dirs, ino0, dirs0, files. ino ∈ dirs ∧ ((abs(sdt0[str], ino0) * {dirs0} ))(dirs) → ino ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino, dirs, files)))) → ino ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino, dirs, files)) ⊦ ino = ino0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: maplet, inst: dir-heap name: maplet seq: ⊦ ((ino => ad))(dirs) ↔ dirs = (∅ ++ ino)[ino, ad] ∧ ino ≠ 0 subst: [dirs, ad, ino] → [dirs2, mkdir(n0, md, m0, n, entries0), ino0] flags: <all paths>

Additional information

Used simplifier rules: ino ≠ 0 ⊦ (∅ ++ ino)[ino, ad] ⊥ dirs ↔ ¬ ino ∈ dirs; (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)% ⊦ ¬ ino ∈ ∅; (s)%ino ≠ 0, ¬ ino ∈ dirs ⊦ ino0 ∈ dirs ++ ino ↔ ino0 = ino ∨ ino0 ∈ dirs; (s) Used terms: dirs2, mkdir(n0, md, m0, n, entries0), ino0