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Proof step 13 from proof abs-path-h (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 14:
((((abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * ino => mkdir(# entries, md, m0, m1, entries)) * D))(dirs), str ∈ entries, str ∈ sdt, ((abs(sdt[str], entries[str]) * D * ino => mkdir(# entries, md, m0, m1, entries) * abs(sdt -- str, entries -- str)))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ sdt[str].fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (ispath(p0, entries[str], dirs, files) ↔ p0 ∈ sdt[str]), ∀ ino0. ⌜ino0⌝ ∈ sdt.fids → ino0 ∈ files, ∀ tree, ino, D, dirs, files. ((abs(tree, ino) * D))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (ispath(p0, ino, dirs, files) ↔ p0 ∈ tree), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ispath(p0, entries[str], dirs, files) ↔ p0 ∈ sdt[str]

Rule argument

pos: left 5 terms: sdt[str], entries[str], D * ino => mkdir(# entries, md, m0, m1, entries) * abs(sdt -- str, entries -- str), dirs, files abort: F discard: F computed: T

Additional information

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