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Proof step 22 from proof ismin (previous step)

b < t .leftmost, b ∈ t, bst(t), t .leftmost ∈ t .left ∨ t .leftmost = t .val, ∀ t0, b. # t0 < # t → (bst(t0) ∧ b ∈ t0 ∧ b < t0 .leftmost → b = t0 .leftmost ∨ t0 = leaf) ⊦ t = leaf, b = t .leftmost

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
t = branch(t1, a, t2), b < t .leftmost, b ∈ t, bst(t), t .leftmost ∈ t1 ∨ t .leftmost = a, ∀ t0, b. # t0 < # t → (bst(t0) ∧ b ∈ t0 ∧ b < t0 .leftmost → b = t0 .leftmost ∨ t0 = leaf) ⊦ t = leaf, b = t .leftmost

Rule argument

[x0lemmaarg: spec-name: tree instance-name: lemma-name: elim antecedent: succedent: t ≠ leaf → ( t0 = t .left ∧ a = t .val ∧ t1 = t .right ↔ t = branch(t0, a, t1)) substlist: [t, t0, a, t1] → [t, t1, a, t2] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ t ≠ leaf → ( t0 = t .left ∧ a = t .val ∧ t1 = t .right ↔ t = branch(t0, a, t1))])], [EXTRATERMS : (t, t1, a, t2)]]