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Proof step 26 from proof split-put (previous step)

str ∈ sdt, p0 ∈ sdt[str], ∀ tree, q, tree0. p0 ∈ tree → tree[p0 + q, tree0] = tree[p0, tree[p0][q, tree0]] ⊦ dnode(md, sdt[str, sdt[str][p0 + q, tree0]]) = dnode(md, sdt[str, sdt[str][p0, sdt[str][p0][q, tree0]]]), p0 = ε

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: sdt[str], q, tree0 abort: F discard: T computed: F

Additional information

Used terms: sdt[str], q, tree0