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Proof step 7 from proof ismin (previous step)

bst(t2), b < a, b ∈ t2, ∀ a0. a0 ∈ t2 → a < a0, ∀ t0, b. # t0 < # branch(leaf, a, t2) → (bst(t0) ∧ b ∈ t0 ∧ b < t0 .leftmost → b = t0 .leftmost ∨ t0 = leaf) ⊦ b = a

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 4, [QUANTTERMLIST : (b)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ a < b ∧ b < c → a < c], [CSIMPSEQ : ⊦ ¬ a < a])], [EXTRATERMS : (b)]]