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Proof step 58 from proof abs-has-rootpath-h (previous step)

((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), str0 ∈ sdi, ino ∈ dirs2, str0 ∈ sdt0, 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))), ∀ str0. str0 ∈ sdi[str, ino1] → str0 ∈ entries0 ∧ sdi[str, ino1][str0] = entries0[str0], ∀ 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)) ⊦ str ∈ sdt0, str ∈ sdi, ino ∈ dirs0, ∃ p. ispath(p, ino0, ino, dirs1 ∪ dirs2, files), str0 ∈ entries0 ∧ sdi[str0] = entries0[str0]

User applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: str0 abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ str ∈ entries ∧ ¬ str0 ∈ entries → entries[str0, ino][str] = entries[str]; (s)% ⊦ str ∈ entries[str0, ino] ↔ str = str0 ∨ str ∈ entries; (ws)%precise(D) ⊦ D(dirs) ∧ dirs ⊥ dirs0 → (((D * D0))(dirs ∪ dirs0) ↔ D0(dirs0)); (s)%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap] Used terms: str0