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Proof step 11 from proof challenge3 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(r => mknode(r1, a, r2) * tr(r1, t0) * tr(r2, t1) * P)(H), bst(t1), bst(t0), (true * tr(r0, branch(t0 .butleftmost, a, t1)) * P)(H0), ∀ a0. ¬ a0 ∈ t0 ∨ ¬ a0 ∈ t1, ∀ a, c. a ∈ t0 ∧ c ∈ t1 → a < c, ∀ a0. a0 ∈ t1 → a < a0, ∀ a0. a0 ∈ t0 → a0 < a ⊦ r = null, t0 = leaf, (a0 ∈ t0 .butleftmost → a0 < a) ∧ bst(t0 .butleftmost)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ P0 * P1 = P1 * P0], [CSIMPSEQ : ⊦ (P0 * P1) * P2 = P0 * P1 * P2])]]