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Proof step 2 from proof abs-fold-new (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
tree[p] = dnode(md, sdt), abs(tree0, ino1) = emp, ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(# entries + 1, md, m0, m1, entries[str, ino1]) * abs(sdt, entries)))(dirs), p ∈ tree ⊦ abs(tree[p + str ', tree0], ROOT_INO)(dirs), str ∈ entries

Proof step 2 from proof abs-fold-new (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information