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Proof step 13 from proof ismin (previous step)

t1 .leftmost ∈ t1, b < t1 .leftmost, bst(t2), bst(t1), b ∈ t1, ∀ a, c. a ∈ t1 ∧ c ∈ t2 → a < c, ∀ a0. ¬ a0 ∈ t1 ∨ ¬ a0 ∈ t2, ∀ a0. a0 ∈ t2 → a < a0, ∀ a0. a0 ∈ t1 → a0 < a, ∀ t0, b. # t0 < # branch(t1, a, t2) → (bst(t0) ∧ b ∈ t0 ∧ b < t0 .leftmost → b = t0 .leftmost ∨ t0 = leaf) ⊦ t1 = leaf, b = t1 .leftmost

User applied rule apply induction

And closed the goal!

Rule argument

[substlist-arg: [t0, b] → [t1, b]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # branch(t, a, t0) = (# t + # t0) +1], [CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ ¬ m + n < m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (t1, b)]]