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Proof step 15 from proof abs-fold-del-p (previous step)

# entries = m + 1, tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(m, md, m0, m1, entries -- str) * abs(sdt -- str, entries -- str) * D))(dirs), p ∈ tree, str ∈ entries ⊦ entries = ∅, ∃ ad. ad.meta = md ∧ ad.size = # ad.entries ∧ ((((ino0 => ad * abs(sdt -- str, ad.entries)) * abs|(tree[p, dnode(ad.meta, sdt -- str)], ROOT_INO, p, ino0)) * D))(dirs)

User applied rule exists right

And obtained the following new goals

Proof step 16:
# entries = m + 1, tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(m, md, m0, m1, entries -- str) * abs(sdt -- str, entries -- str) * D))(dirs), str ∈ entries, p ∈ tree ⊦ entries = ∅, mkdir(m, md, m0, m1, entries -- str).meta = md ∧ mkdir(m, md, m0, m1, entries -- str).size = # mkdir(m, md, m0, m1, entries -- str).entries ∧ (( ( (ino0 => mkdir(m, md, m0, m1, entries -- str) * abs(sdt -- str, mkdir(m, md, m0, m1, entries -- str).entries)) * abs|(tree[p, dnode(mkdir(m, md, m0, m1, entries -- str).meta, sdt -- str)], ROOT_INO, p, ino0)) * D) )(dirs)

Rule argument

pos: right 2 terms: mkdir(m, md, m0, m1, entries -- str) abort: F discard: T computed: T

Additional information

Used terms: mkdir(m, md, m0, m1, entries -- str)