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Proof step 16 from proof tree-ext (previous step)

dnode(md, sdt) = tree1, ∀ p. (p ∈ dnode(md, sdt) ↔ p ∈ tree2) ∧ ( p ∈ dnode(md, sdt) → (dnode(md, sdt)[p].dir? ↔ tree2[p].dir?) ∧ (dnode(md, sdt)[p].dir? ⊃ dnode(md, sdt)[p].meta = tree2[p].meta;dnode(md, sdt)[p].fid = tree2[p].fid)), ∀ str. str ∈ sdt → (∀ tree2. (∀ p. (p ∈ sdt[str] ↔ p ∈ tree2) ∧ ( p ∈ sdt[str] → (sdt[str][p].dir? ↔ tree2[p].dir?) ∧ (sdt[str][p].dir? ⊃ sdt[str][p].meta = tree2[p].meta;sdt[str][p].fid = tree2[p].fid))) → sdt[str] = tree2) ⊦ dnode(md, sdt) = tree2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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