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Proof step 58 from proof in-put (previous step)

dnode(md1, sdt2)[p0] = dnode(md, sdt), p0 ∈ dnode(md1, sdt2), ∀ 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])] ↔ (p0 + str ' ⊑ q ⊃ restn(# p0, q.tail) ∈ tree0;q ∈ dnode(md0, sdt0))) ⊦ q0 ∈ dnode(md1, sdt2)[p0, dnode(md, sdt[str, tree0])] ↔ (p0 + str ' ⊑ q0 ⊃ restn(# p0, q0.tail) ∈ tree0;q0 ∈ dnode(md1, sdt2)), p0 = ε, str0 ∈ sdt1

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: md1, sdt2, md, sdt, q0, str, tree0 abort: F discard: F computed: T

Additional information

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