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Proof step 31 from proof at-del (previous step)

str ∈ sdt, p0 ∈ sdt[str], q1 ∈ sdt[str], ∀ tree, q. p0 ∈ tree ∧ ¬ p0 ⊑ q ∧ q ∈ tree → (tree -- p0)[q] = (q ⊑ p0 ⊃ tree[q] -- restn(# q, p0);tree[q]) ⊦ (sdt[str] -- p0)[q1] = (q1 ⊑ p0 ⊃ sdt[str][q1] -- restn(# q1, p0);sdt[str][q1]), p0 = ε, p0 ⊑ q1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: sdt[str], q1

Additional information

Used terms: sdt[str], q1