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Proof step 49 from proof challenge3 (previous step)

bst(t1), bst(t0), (true * tr(r0, (t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))) * P)(H0), (r => mknode(r1, a, r2) * tr(r1, t0) * tr(r2, t1) * P)(H), b ∈ ((t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))), ∀ 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 ⊃ a; t0 .leftmost)) < b

User applied rule case distinction

And obtained the following new goals

Proof step 35:
t0 = leaf, bst(t1), bst(t0), (true * tr(r0, (t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))) * P)(H0), (r => mknode(r1, a, r2) * tr(r1, t0) * tr(r2, t1) * P)(H), b ∈ ((t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))), ∀ a0. ¬ a0 ∈ t0 ∨ ¬ a0 ∈ t1, ∀ a, c. a ∈ t0 ∧ c ∈ t1 → a < c, ∀ a0. a0 ∈ t1 → a < a0, ∀ a0. a0 ∈ t0 → a0 < a ⊦ a < b, r = null
Proof step 48:
t0 ≠ leaf, bst(t1), bst(t0), (true * tr(r0, (t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))) * P)(H0), (r => mknode(r1, a, r2) * tr(r1, t0) * tr(r2, t1) * P)(H), b ∈ ((t0 = leaf ⊃ t1; branch(t0 .butleftmost, a, t1))), ∀ a0. ¬ a0 ∈ t0 ∨ ¬ a0 ∈ t1, ∀ a, c. a ∈ t0 ∧ c ∈ t1 → a < c, ∀ a0. a0 ∈ t1 → a < a0, ∀ a0. a0 ∈ t0 → a0 < a ⊦ t0 .leftmost < b, r = null

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]