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Proof step 23 from proof abs-has-rootpath-h (previous step)

ino ∈ dirs, ∃ dirs2, dirs1. dirs2 ⊥ dirs1 ∧ dirs = dirs2 ∪ dirs1 ∧ ((ino0 => mkdir(n0, md, m0, n, entries0)))(dirs2) ∧ {dirs0} (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, ino ∈ dirs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
((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

Rule argument

Additional information

Used simplifier rules: ⊦ ino ∈ dirs ∧ dirs ⊥ dirs0 → (ino ∈ dirs0 ↔ false); (s)% ⊦ dirs ⊥ dirs0 → (ino ∈ dirs ∪ dirs0 ↔ ino ∈ dirs ∨ ino ∈ dirs0); (s)% ⊦ {dirs} (dirs0) ↔ dirs = dirs0; (s)