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Proof step 2 from proof dir-put (previous step)

tree[p] = dnode(md, sdt), tree.dir?, p ∈ tree ⊦ tree[p, dnode(md, sdt[str, tree0])].dir?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ tree.dir? ↔ (∃ md, sdt. tree = dnode(md, sdt))
Proof step 4:
⊦ tree.dir? ↔ ¬(∃ fid. tree = fnode(fid))
Proof step 5:
⊦ p ∈ tree → (tree[p, tree0] = fnode(fid) ↔ p = ε ∧ tree0 = fnode(fid))

Rule argument

Additional information

Used simplifier rules: dnode-elim-strong: ⊦ tree.dir? ↔ (∃ md, sdt. tree = dnode(md, sdt));%fnode-elim-strong: ⊦ tree.dir? ↔ ¬(∃ fid. tree = fnode(fid));%put-fnode: ⊦ p ∈ tree → (tree[p, tree0] = fnode(fid) ↔ p = ε ∧ tree0 = fnode(fid));% ⊦ fnode(fid) ≠ dnode(md, sdt); (s)