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Proof step 37 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(tree, ino1)(dirs1), ispath(p, ino1, ino, dirs1, files), ino ∈ dirs1, sdi[str, ino1] ⊆ entries0, dirs1 ⊥ dirs2, ∀ 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, dirs1 ∪ dirs2, files)

User applied rule exists right

And obtained the following new goals

Proof step 38:
((abs(tree, ino1) * abs(sdt0, sdi) * {dirs0} * ino0 => mkdir(n0, md, m0, n, entries0)))(dirs1 ∪ dirs2), ((ino0 => mkdir(n0, md, m0, n, entries0) * {dirs0} * abs(sdt0, sdi)))(dirs2), abs(tree, ino1)(dirs1), ispath(p, ino1, ino, dirs1, files), ino ∈ dirs1, sdi[str, ino1] ⊆ entries0, dirs1 ⊥ dirs2, ∀ 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)) ⊦ ispath(str + p, ino0, ino, dirs1 ∪ dirs2, files), str ∈ sdt0, str ∈ sdi, ino ∈ dirs0

Rule argument

pos: right 4 terms: str + p abort: F discard: T computed: F

Additional information

Used terms: str + p