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Proof step 27 from proof absc-drop (previous step)

(( ((abs|(sdt[str0][p0, tree0], entries[str0], q0, ino1) * abs(sdt -- str0, entries -- str0)) * ino0 => mkdir(# entries, md, m0, m1, entries)) * D) )(dirs), sdt[str0][parent(p0)].dir?, str0 ∈ entries, str0 ∈ sdt, parent(p0) ∈ sdt[str0], q0 ⊑ p0, ∀ tree, p, tree0, ino0, ino1, D, dirs. tree[parent(p)].dir? ∧ parent(p) ∈ tree ∧ q0 ⊑ p → (((abs|(tree[p, tree0], ino0, q0, ino1) * D))(dirs) ↔ ((abs|(tree, ino0, q0, ino1) * D))(dirs)) ∨ p = ε ⊦ p0 = ε, ∃ ad0. str0 ∈ ad0.entries ∧ ad0.meta = md ∧ ad0.size = # ad0.entries ∧ ((((abs|(sdt[str0], ad0.entries[str0], q0, ino1) * abs(sdt -- str0, ad0.entries -- str0)) * ino0 => ad0) * D))(dirs)

User applied rule all left

And obtained the following new goals

Proof step 28:
(( ((abs|(sdt[str0][p0, tree0], entries[str0], q0, ino1) * abs(sdt -- str0, entries -- str0)) * ino0 => mkdir(# entries, md, m0, m1, entries)) * D) )(dirs), sdt[str0][parent(p0)].dir?, parent(p0) ∈ sdt[str0], str0 ∈ sdt, str0 ∈ entries, q0 ⊑ p0, sdt[str0][parent(p0)].dir? ∧ parent(p0) ∈ sdt[str0] ∧ q0 ⊑ p0 → ( (( abs|(sdt[str0][p0, tree0], entries[str0], q0, ino1) * D * ino0 => mkdir(# entries, md, m0, m1, entries) * abs(sdt -- str0, entries -- str0)) )(dirs) ↔ ((abs|(sdt[str0], entries[str0], q0, ino1) * D * ino0 => mkdir(# entries, md, m0, m1, entries) * abs(sdt -- str0, entries -- str0)) )(dirs)) ∨ p0 = ε ⊦ p0 = ε, ∃ ad0. str0 ∈ ad0.entries ∧ ad0.meta = md ∧ ad0.size = # ad0.entries ∧ ((((abs|(sdt[str0], ad0.entries[str0], q0, ino1) * abs(sdt -- str0, ad0.entries -- str0)) * ino0 => ad0) * D))(dirs)

Rule argument

pos: left 7 terms: sdt[str0], p0, tree0, entries[str0], ino1, D * ino0 => mkdir(# entries, md, m0, m1, entries) * abs(sdt -- str0, entries -- str0), dirs abort: F discard: T computed: T

Additional information

Used terms: sdt[str0], p0, tree0, entries[str0], ino1, D * ino0 => mkdir(# entries, md, m0, m1, entries) * abs(sdt -- str0, entries -- str0), dirs