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Proof step 5 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) ⊦ abs(tree[p, tree[p][str ', tree0]], ROOT_INO)(dirs), str ∈ entries

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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)(dirs), str ∈ entries

Rule argument

Additional information

Used simplifier rules: ⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0]); (s)