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Proof step 14 from proof abs-has-rootpath-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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