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Proof step 62 from proof at-mod (previous step)

tree[p0] = dnode(md0, sdt0), tree = dnode(md2, sdt3), q0 ∈ tree[p0, dnode(md1, sdt1)], p0 ∈ tree, ∀ md, sdt, md0, sdt0, q, md1, sdt1. dnode(md, sdt)[p0] = dnode(md0, sdt0) ∧ p0 ∈ dnode(md, sdt) ∧ q ∈ dnode(md, sdt)[p0, dnode(md1, sdt1)] → dnode(md, sdt)[p0, dnode(md1, sdt1)][q] = (q ⊑ p0 ⊃ dnode(md, sdt)[q][restn(# q, p0), dnode(md1, sdt1)];dnode(md, sdt)[p0, dnode(md0, sdt1)][q]) ⊦ tree[p0, dnode(md1, sdt1)][q0] = (q0 ⊑ p0 ⊃ tree[q0][restn(# q0, p0), dnode(md1, sdt1)];tree[p0, dnode(md0, sdt1)][q0]), p0 = ε, str ∈ sdt2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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