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Proof step 9 from proof abs-fold-new-p (previous step)

tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(# entries + 1, md, m0, m1, entries[str, ino1]) * abs(sdt, entries) * D))(dirs), p ∈ tree, D ⊆ abs(tree0, ino1) ⊦ str ∈ entries, ∃ ino. ((abs|(tree[p, dnode(md, sdt[str, tree0])], ROOT_INO, p, ino) * abs(dnode(md, sdt[str, tree0]), ino)))(dirs)

User applied rule exists right

And obtained the following new goals

Proof step 10:
tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(# entries + 1, md, m0, m1, entries[str, ino1]) * abs(sdt, entries) * D))(dirs), p ∈ tree, D ⊆ abs(tree0, ino1) ⊦ ((abs|(tree[p, dnode(md, sdt[str, tree0])], ROOT_INO, p, ino0) * abs(dnode(md, sdt[str, tree0]), ino0)))(dirs), str ∈ entries

Rule argument

pos: right 2 terms: ino0 abort: F discard: T computed: F

Additional information

Used terms: ino0