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Proof step 18 from proof challenge3 (previous step)

(true * tr(r0, t1) * P)(H0), ((tr(r2, t1) * r => mknode(null, a, r2)) * P)(H), a ∈ t1, bst(t1), ∀ a0. a0 ∈ t1 → a < a0 ⊦ r = null

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 5, [QUANTTERMLIST : (a)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ a < a])], [EXTRATERMS : (a)]]