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Proof step 25 from proof challenge3 (previous step)

bst(t0), bst(t1), (r => mknode(r1, a, r2) * tr(r1, t0) * tr(r2, t1) * P)(H), (true * tr(r0, branch(t0 .butleftmost, a, t1)) * P)(H0), t0 .leftmost < a, t0 .leftmost ∈ t1, ∀ a0. a0 ∈ t0 → a0 < a, ∀ a0. a0 ∈ t1 → a < a0, ∀ a, c. a ∈ t0 ∧ c ∈ t1 → a < c, ∀ a0. ¬ a0 ∈ t0 ∨ ¬ a0 ∈ t1 ⊦ r = null, t0 = leaf, t0 .leftmost = a

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 8, [QUANTTERMLIST : (t0 .leftmost)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (P0 * P1) * P2 = P0 * P1 * P2], [CSIMPSEQ : ⊦ P0 * P1 = P1 * P0], [CSIMPFORWARD : ⊦ a < b ∧ b < c → a < c], [CSIMPSEQ : ⊦ ¬ a < a])], [EXTRATERMS : (t0 .leftmost)]]