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Proof step 46 from proof abs-has-rootpath-h (previous step)

((abs(tree, ino1) * abs(sdt0, sdi) * {dirs0} * ino0 => mkdir(n0, md, m0, n, entries1)))(dirs1 ∪ dirs2), ((ino0 => mkdir(n0, md, m0, n, entries1) * {dirs0} * abs(sdt0, sdi)))(dirs2), abs(tree, ino1)(dirs1), ispath(p, ino1, ino, dirs1, files), ino ∈ dirs1, dirs1 ⊥ dirs2, ∀ str0. str0 ∈ sdi[str, ino1] → str0 ∈ entries1 ∧ sdi[str, ino1][str0] = entries1[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)) ⊦ ino0 ∈ dirs1, str ∈ sdt0, str ∈ sdi, ino ∈ dirs0, str ∈ entries1 ∧ ispath(p, entries1[str], ino, dirs1 ∪ dirs2, files)

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: str abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ str ∈ entries[str, ino]; (s)% ⊦ entries[str, ino][str] = ino; (s)%⊦ dirs0 ⊥ dirs1 ∧ ispath(p, n0, n1, dirs0, files) → ispath(p, n0, n1, dirs0 ∪ dirs1, files); (ss)%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: str