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Proof step 51 from proof challenge3 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 15:
¬ bst((t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))), (r => mknode(r1, a, r2) * tr(r1, t0) * tr(r2, t1) * P)(H), (true * tr(r0, (t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))) * P)(H0), bst(t0), bst(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
Proof step 31:
((t0 = leaf ⊃ a; t0 .leftmost)) ∈ ((t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))), (r => mknode(r1, a, r2) * tr(r1, t0) * tr(r2, t1) * P)(H), (true * tr(r0, (t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))) * P)(H0), bst(t0), bst(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
Proof step 50:
¬ (b ∈ ((t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))) → ((t0 = leaf ⊃ a; t0 .leftmost)) < b), (r => mknode(r1, a, r2) * tr(r1, t0) * tr(r2, t1) * P)(H), (true * tr(r0, (t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))) * P)(H0), bst(t0), bst(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

Rule argument

[CASEDARG : pos: left 8, ()]

Additional information

[[CONCRETESIMPRULES : ()]]