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Proof step 18 from proof same-put (
previous step
)
str ∈ sdt, p0 ∈ sdt[str], ∀ tree, tree0, tree1. p0 ∈ tree → tree[p0, tree0][p0, tree1] = tree[p0, tree1] ⊦ dnode(md, sdt[str, sdt[str][p0, tree0][p0, tree1]]) = dnode(md, sdt[str, sdt[str][p0, tree1]]), p0 = ε
User applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: sdt[str], tree0, tree1 abort: F discard: T computed: F
Additional information
Used terms: sdt[str], tree0, tree1