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Proof step 64 from proof abs-has-rootpath-h (previous step)

((ino0 => mkdir(n0, md, m0, n, entries0) * {dirs0} * abs(sdt0, sdi)))(dirs2), ((abs(tree, ino1) * abs(sdt0, sdi) * {dirs0} * ino0 => mkdir(n0, md, m0, n, entries0)))(dirs1 ∪ dirs2), ((abs(sdt0[str0], ino3) * {dirs3} ))(dirs), abs(tree, ino1)(dirs1), str0 ∈ sdt0, ino ∈ dirs2, ino2 ∈ dirs, str0 ∈ sdi, str0 ∈ sdt0[str, tree], sdi[str, ino1] ⊆ entries0, dirs1 ⊥ dirs2, ∀ str0. str0 ∈ sdt0[str, tree] → (∀ ino, dirs, ino0, dirs0, files. ino ∈ dirs ∧ ((abs(sdt0[str, tree][str0], ino0) * {dirs0} ))(dirs) → ino ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino, dirs, files))), ∀ ino4, dirs4, ino5, dirs5, files1. ino4 ∈ dirs4 ∧ ((abs(sdt0[str0], ino5) * {dirs5} ))(dirs4) → ino4 ∈ dirs5 ∨ (∃ p. ispath(p, ino5, ino4, dirs4, files1)), ∀ sdt1, ino0, n0, md, m0, n, entries0, entries, dirs0, dirs, ino, files. # sdt1 < # sdt0[str, tree] → (((ino0 => mkdir(n0, md, m0, n, entries0) * abs(sdt1, entries)) * {dirs0} ))(dirs) ∧ ino ∈ dirs ∧ entries ⊆ entries0 ∧ (∀ str. str ∈ sdt1 → (∀ ino, dirs, ino0, dirs0, files. ino ∈ dirs ∧ ((abs(sdt1[str], ino0) * {dirs0} ))(dirs) → ino ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino, dirs, files)))) → ino ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino, dirs, files)) ⊦ ino2 ∈ dirs3, ino ∈ dirs0, str ∈ sdi, str ∈ sdt0, ∃ p. ispath(p, ino3, ino2, dirs, files0), ∃ p. ispath(p, ino0, ino, dirs1 ∪ dirs2, files)

Proof step 64 from proof abs-has-rootpath-h (previous step)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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