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Proof step 47 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, t0 .leftmost < b, b ≠ a ∧ ¬ b ∈ t0 .butleftmost ∧ ¬ b ∈ t1

User applied rule case distinction

And obtained the following new goals

Proof step 39:
(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 ⊦ b ≠ a, r = null, t0 = leaf, t0 .leftmost < b
Proof step 43:
(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 ⊦ ¬ b ∈ t0 .butleftmost, r = null, t0 = leaf, t0 .leftmost < b
Proof step 46:
(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 ⊦ ¬ b ∈ t1, r = null, t0 = leaf, t0 .leftmost < b

Rule argument

[CASEDARG : pos: right 4, ()]

Additional information

[[CONCRETESIMPRULES : ()]]