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Proof step 87 from proof at-put (previous step)

dnode(md1, sdt2)[p0] = dnode(md, sdt), p0 ∈ dnode(md1, sdt2), q0 ∈ dnode(md1, sdt2)[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])) ⊦ dnode(md1, sdt2)[p0, dnode(md, sdt[str, tree0])][q0] = (p0 + str ' ⊑ q0 ⊃ tree0[restn(# p0, q0.tail)];(q0 ⊑ p0 ⊃ dnode(md1, sdt2)[q0][restn(# q0, p0) + str ', tree0];dnode(md1, sdt2)[q0])), p0 = ε, str0 ∈ sdt1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: md1, sdt2, md, sdt, q0, str, tree0

Additional information

Used terms: md1, sdt2, md, sdt, q0, str, tree0