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Proof step 21 from proof abs-dir-h (previous step)

((ino => mkdir(# entries, md, m0, m1, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * D))(dirs), ispath(p0, entries[str], ino0, dirs, files), str ∈ sdt, str ∈ entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ tree, ino, D, dirs, ino0, files. ((abs(tree, ino) * D))(dirs) ∧ ispath(p0, ino, ino0, dirs, files) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (tree[p0].dir? ↔ ino0 ∈ dirs), ∀ ino1. ⌜ino1⌝ ∈ sdt.fids → ino1 ∈ files ⊦ sdt[str][p0].dir? ↔ ino0 ∈ dirs

User applied rule all left

And obtained the following new goals

Proof step 22:
((ino => mkdir(# entries, md, m0, m1, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * D))(dirs), ispath(p0, entries[str], ino0, dirs, files), str ∈ entries, str ∈ sdt, ((abs(sdt[str], entries[str]) * D * abs(sdt -- str, entries -- str) * ino => mkdir(# entries, md, m0, m1, entries)))(dirs) ∧ ispath(p0, entries[str], ino0, dirs, files) ∧ (∀ ino. ⌜ino⌝ ∈ sdt[str].fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (sdt[str][p0].dir? ↔ ino0 ∈ dirs), ∀ ino1. ⌜ino1⌝ ∈ sdt.fids → ino1 ∈ files, ∀ tree, ino, D, dirs, ino0, files. ((abs(tree, ino) * D))(dirs) ∧ ispath(p0, ino, ino0, dirs, files) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (tree[p0].dir? ↔ ino0 ∈ dirs), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ sdt[str][p0].dir? ↔ ino0 ∈ dirs

Rule argument

pos: left 6 terms: sdt[str], entries[str], D * abs(sdt -- str, entries -- str) * ino => mkdir(# entries, md, m0, m1, entries), dirs, ino0, files abort: F discard: F computed: T

Additional information

Used terms: sdt[str], entries[str], D * abs(sdt -- str, entries -- str) * ino => mkdir(# entries, md, m0, m1, entries), dirs, ino0, files