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Proof step 15 from proof same-in (
previous step
)
str ∈ sdt, p0 ∈ sdt[str], ∀ tree, tree0. p0 ∈ tree → p0 ∈ tree[p0, tree0] ⊦ p0 ∈ sdt[str][p0, tree0]
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: sdt[str], tree0
Additional information
Used terms: sdt[str], tree0