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

p ∈ tree ⊦ (tree -- p).dir? ↔ tree.dir?, p = ε

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ ¬ tree.dir? → (p ∈ tree ↔ p = ε)
Proof step 4:
⊦ tree.dir? ↔ (∃ md, sdt. tree = dnode(md, sdt))
Proof step 5:
⊦ tree.dir? ↔ ¬(∃ fid. tree = fnode(fid))
Proof step 6:
p ≠ ε ⊦ dnode(md, sdt) -- p ≠ fnode(fid)

Rule argument

Additional information

Used simplifier rules: notdir-notin: ⊦ ¬ tree.dir? → (p ∈ tree ↔ p = ε);%dnode-elim-strong: ⊦ tree.dir? ↔ (∃ md, sdt. tree = dnode(md, sdt));%fnode-elim-strong: ⊦ tree.dir? ↔ ¬(∃ fid. tree = fnode(fid));%dnode-del: p ≠ ε ⊦ dnode(md, sdt) -- p ≠ fnode(fid);