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Proof step 84 from proof at-put (previous step)

sdt0[str0][p0] = dnode(md, sdt), q0 ∈ sdt0[str0][p0, dnode(md, sdt[str, tree0])], p0 ∈ sdt0[str0], ∀ md0, sdt0, md, sdt, q, str, tree0. dnode(md0, sdt0)[p0] = dnode(md, sdt) ∧ p0 ∈ dnode(md0, sdt0) ∧ q ∈ dnode(md0, sdt0)[p0, dnode(md, sdt[str, tree0])] → dnode(md0, sdt0)[p0, dnode(md, sdt[str, tree0])][q] = (p0 + str ' ⊑ q ⊃ tree0[restn(# p0, q.tail)];(q ⊑ p0 ⊃ dnode(md0, sdt0)[q][restn(# q, p0) + str ', tree0];dnode(md0, sdt0)[q])), ∃ tree, sdt1. sdt0 = sdt1[str0, tree] ∧ ¬ str0 ∈ sdt1 ⊦ sdt0[str0][p0, dnode(md, sdt[str, tree0])][q0] = (p0 + str ' ⊑ q0 ⊃ tree0[restn(# p0, q0.tail)];(q0 ⊑ p0 ⊃ sdt0[str0][q0][restn(# q0, p0) + str ', tree0];sdt0[str0][q0])), p0 = ε

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 85:
tree[p0] = dnode(md, sdt), p0 ∈ tree, q0 ∈ tree[p0, dnode(md, sdt[str, tree0])], ∀ md0, sdt0, md, sdt, q, str, tree0. dnode(md0, sdt0)[p0] = dnode(md, sdt) ∧ p0 ∈ dnode(md0, sdt0) ∧ q ∈ dnode(md0, sdt0)[p0, dnode(md, sdt[str, tree0])] → dnode(md0, sdt0)[p0, dnode(md, sdt[str, tree0])][q] = (p0 + str ' ⊑ q ⊃ tree0[restn(# p0, q.tail)];(q ⊑ p0 ⊃ dnode(md0, sdt0)[q][restn(# q, p0) + str ', tree0];dnode(md0, sdt0)[q])) ⊦ tree[p0, dnode(md, sdt[str, tree0])][q0] = (p0 + str ' ⊑ q0 ⊃ tree0[restn(# p0, q0.tail)];(q0 ⊑ p0 ⊃ tree[q0][restn(# q0, p0) + str ', tree0];tree[q0])), p0 = ε, str0 ∈ sdt1

Rule argument

Additional information

Used simplifier rules: ⊦ sdt[str, tree][str] = tree; (s)