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Proof step 10 from proof tree-ext (previous step)

⊦ (fnode(fid) = tree ∧ (∀ p. (p ∈ tree0 → p = ε) ∧ (p = ε → ¬ tree0.dir? ∧ fid = tree0.fid)) → fnode(fid) = tree0) ∧ ( (∀ 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) = 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))) → dnode(md, sdt) = tree2)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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