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Proof step 26 from proof abs-has-rootpath-h (previous step)

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))), ∃ tree, ino, str, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ entries = sdi0[str, ino] ∧ ((abs(tree, ino) * abs(sdt0, sdi0) * {dirs0} * ino0 => mkdir(n0, md, m0, n, entries0)))(dirs), ∀ 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)) ⊦ sdt = ∅, ino ∈ dirs0, ∃ p. ispath(p, ino0, ino, dirs, files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
((abs(tree, ino1) * abs(sdt0, sdi) * {dirs0} * ino0 => mkdir(n0, md, m0, n, entries0)))(dirs), ino ∈ dirs, sdi[str, ino1] ⊆ entries0, ∀ 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))), ∀ 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)) ⊦ ino ∈ dirs0, str ∈ sdi, str ∈ sdt0, ∃ p. ispath(p, ino0, ino, dirs, files)

Rule argument

Additional information

Used simplifier rules: ⊦ sdt[str, tree] ≠ ∅; (s)